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For this Question: Andy loves Xor. Dwight gives Andy an integer N and asks him to tell whether XOR of all the integers from 1 to N is even or odd. As Andy is busy flirting with Erin, help him solve this problem.Input contains a single integer N.
Constraints:
1 <= N <= 10^9Print "even" if XOR of all the integers from 1 to N is even and if it is odd print "odd".Sample Input 1
3
Sample Output 1
even
Explanation: Xor of 1, 2 and 3 is 0 which is even.
Sample Input 2
5
Sample Output 2
odd
Explanation: Xor of 1, 2, 3, 4 and 5 is 1 which is odd., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
int input= Integer.parseInt(reader.readLine());
int result=1;
for(int i=2;i<=input;i++)
{
result = result^i;
}
if(result%2 == 0)
System.out.println("even");
else
System.out.println("odd");
}
catch(Exception e){
System.out.println("Something went wrong"+e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andy loves Xor. Dwight gives Andy an integer N and asks him to tell whether XOR of all the integers from 1 to N is even or odd. As Andy is busy flirting with Erin, help him solve this problem.Input contains a single integer N.
Constraints:
1 <= N <= 10^9Print "even" if XOR of all the integers from 1 to N is even and if it is odd print "odd".Sample Input 1
3
Sample Output 1
even
Explanation: Xor of 1, 2 and 3 is 0 which is even.
Sample Input 2
5
Sample Output 2
odd
Explanation: Xor of 1, 2, 3, 4 and 5 is 1 which is odd., I have written this Solution Code: n=int(input())
odds = n//2 if n%2==0 else (n+1)//2
if(odds % 2 == 0 ):
print("even")
else:
print("odd"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Andy loves Xor. Dwight gives Andy an integer N and asks him to tell whether XOR of all the integers from 1 to N is even or odd. As Andy is busy flirting with Erin, help him solve this problem.Input contains a single integer N.
Constraints:
1 <= N <= 10^9Print "even" if XOR of all the integers from 1 to N is even and if it is odd print "odd".Sample Input 1
3
Sample Output 1
even
Explanation: Xor of 1, 2 and 3 is 0 which is even.
Sample Input 2
5
Sample Output 2
odd
Explanation: Xor of 1, 2, 3, 4 and 5 is 1 which is odd., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%4==1||n%4==2)
cout<<"odd";
else
cout<<"even";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in =new BufferedReader(new InputStreamReader(System.in));
StringBuilder s = new StringBuilder();
String text=null;
while ((text = in.readLine ()) != null)
{
s.append(text);
}
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)==s.charAt(i+1)){
int flag=0;
s.delete(i,i+2);
int left=i-1;
len=len-2;
i=i-2;
if(i<0){
i=-1;
}
}
}
System.out.println(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: s=input()
l=["aa","bb","cc","dd","ee","ff","gg","hh","ii","jj","kk","ll","mm","nn","oo","pp","qq","rr","ss","tt","uu","vv","ww","xx","yy","zz"]
while True:
do=False
for i in range(len(l)):
if l[i] in s:
do=True
while l[i] in s:
s=s.replace(l[i],"")
if do==False:
break
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int len=s.length();
char stk[410000];
int k = 0;
for (int i = 0; i < len; i++)
{
stk[k++] = s[i];
while (k > 1 && stk[k - 1] == stk[k - 2])
k -= 2;
}
for (int i = 0; i < k; i++)
cout << stk[i];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.HashMap;
import java.util.Map;
public class Main {
public static void main(String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String line = br.readLine();
int size = Integer.parseInt(line);
String str = br.readLine();
String[] strArray = str.split(" ");
int[] array = new int[size];
for (int i = -0; i < size; i++) {
array[i] = Integer.parseInt(strArray[i]);
}
int count = largestSubarray(array,size);
System.out.println(count);
}
static int largestSubarray(int[] array,int size){
int count = -1;
int sum = 0;
Map<Integer,Integer> mymap = new HashMap<>();
mymap.put(0,-1);
for(int i=0; i<array.length; i++){
sum += array[i];
if(mymap.containsKey(sum)){
count = Math.max(count, i-mymap.get(sum));
}
else{
mymap.put(sum,i);
}
}
if(count > 0){
return count;
}
else{
return -1;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: def maxLen(arr,n,k):
mydict = dict()
sum = 0
maxLen = 0
for i in range(n):
sum += arr[i]
if (sum == k):
maxLen = i + 1
elif (sum - k) in mydict:
maxLen = max(maxLen, i - mydict[sum - k])
if sum not in mydict:
mydict[sum] = i
return maxLen
n=int(input())
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
max_len=maxLen(arr,n,0)
if(max_len==0):
print ("-1")
else:
print (max_len), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>A[]</b>, of length <b>N</b> containing zero , positive and negative integers.
You need to find the length of the largest subarray whose sum of elements is <b>0</b>.The first line contains N denoting the size of the array A. Then in the next line contains N space-separated values of the array A.
<b>Constraints:-</b>
1 <= N <= 1e5
-1e6 <= A[i] <= 1e6Print the length of the largest subarray which has sum 0, If no subarray exist print -1.Sample Input:-
8
15 -2 2 -8 1 7 10 23
Sample Output:-
5
Explanation:-
-2 2 -8 1 7 is the required subarray
Sample Input:-
5
1 2 1 2 3
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unordered_map<long long,int> m;
int n,k;
cin>>n;
long a[n];
int ans=-1;
for(int i=0;i<n;i++){cin>>a[i];if(a[i]==0){ans=1;}}
long long sum=0;
for(int i=0;i<n;i++){
sum+=a[i];
if(sum==0){ans=max(i+1,ans);}
if(m.find(sum)==m.end()){m[sum]=i;}
else{
ans=max(i-m[sum],ans);
}
}
cout<<ans;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code: public static long SumOfDivisors(long N){
long sum=0;
long c=(long)Math.sqrt(N);
for(long i=1;i<=c;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){sum+=N/i;}
}
}
return sum;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
long long SumOfDivisors(long long N){
long long sum=0;
long sq=sqrt(N);
for(long i=1;i<=sq;i++){
if(N%i==0){
sum+=i;
if(i*i!=N){
sum+=N/i;
}
}
}
return sum;
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, your task is to return the sum of all of its divisors.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SumOfDivisors()</b> that takes the integer N as parameter.
Constraints:-
1<=N<=10^9Return the sum of all of the divisors.Sample Input:-
4
Sample Output:-
7
Sample Input:-
13
Sample Output:-
14, I have written this Solution Code:
def SumOfDivisors(num) :
# Final result of summation of divisors
result = 0
# find all divisors which divides 'num'
i = 1
while i<= (math.sqrt(num)) :
# if 'i' is divisor of 'num'
if (num % i == 0) :
# if both divisors are same then
# add it only once else add both
if (i == (num / i)) :
result = result + i;
else :
result = result + (i + num/i);
i = i + 1
# Add 1 to the result as 1 is also
# a divisor
return (result); , In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define N 1000
// Function to rotate the matrix 90 degree clockwise
void rotate90Clockwise(int a[][N],int n)
{
// Traverse each cycle
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
// Swap elements of each cycle
// in clockwise direction
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
}
// Function for print matrix
void printMatrix(int arr[][N],int n)
{
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++)
cout << arr[i][j] << " ";
cout << '\n';
}
}
// Driver code
int main()
{
int n;
cin>>n;
int arr[n][N];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
cin>>arr[i][j];
}}
rotate90Clockwise(arr,n);
printMatrix(arr,n);
cout<<endl;
rotate90Clockwise(arr,n);
printMatrix(arr,n);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code:
import java.util.*;
import java.lang.*;
import java.io.*;
class Main
{
public static void main (String[] args) throws java.lang.Exception
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int a[][] = new int[n][n];
for(int i=0;i<n;i++){
for(int j=0;j<n;j++){
a[i][j]= sc.nextInt();
}
}
//rotating by 90 degree
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
for(int i =0;i<n;i++){
for(int j=0;j<n;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
//rotating by 90 degree
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - i - 1; j++) {
int temp = a[i][j];
a[i][j] = a[n - 1 - j][i];
a[n - 1 - j][i] = a[n - 1 - i][n - 1 - j];
a[n - 1 - i][n - 1 - j] = a[j][n - 1 - i];
a[j][n - 1 - i] = temp;
}
}
System.out.println();
for(int i =0;i<n;i++){
for(int j=0;j<n;j++){
System.out.print(a[i][j]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a matrix of N*N dimensions (Mat). Print the matrix rotated by 90 degrees and 180 degrees.The first line contains N.
N lines follow each containing N space-separated integers.
<b>Constraints</b>
2 <= N <= 100
1 <= Mat[i][j] <= 10000Output 2*N+1 lines.
First N lines should contain the Matrix rotated by 90 degrees.
Then print a blank line.
Then N lines should contain the Matrix rotated by 180 degrees.Sample Input 1:
2
3 4
7 6
Sample Output 1:
7 3
6 4
6 7
4 3
Sample Input 2:
2
1 2
3 4
Sample Output 2:
3 1
4 2
4 3
2 1, I have written this Solution Code: x=int(input())
l1=[]
for i in range(x):
a1=list(map(int,input().split()))
l1.append(a1)
for j in range(x):
for i in range(1,x+1):
print(l1[-i][j], end=" ")
print()
print()
for i in range(1,x+1):
for j in range(1,x+1):
print(l1[-i][-j], end=" ")
print(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader bd=new BufferedReader(new InputStreamReader(System.in));
String st1=bd.readLine();
int n=Integer.parseInt(st1);
int arr[]=new int[n];
String[] st2=bd.readLine().split(" ");
for(int i=0; i<n; i++){
arr[i]=Integer.parseInt(st2[i]);
}
long sum_till_here_forward = 0;
long max_sum_forward = 0;
long sum_till_here_backward = 0;
long max_sum_backward = 0;
long sum_forward[] = new long[n+1];
long sum_backward[] = new long[n+1];
long maximum=0;
if(n==2){
maximum=Math.max(arr[0],arr[1]);
}
else{
for(int i=0; i<n; i++){
sum_till_here_forward += arr[i];
if(sum_till_here_forward > max_sum_forward){
max_sum_forward = sum_till_here_forward;
}
if(sum_till_here_forward < 0){
sum_till_here_forward = 0;
}
sum_forward[i+1] = max_sum_forward;
}
for(int i=n-1; i>=0; i--){
sum_till_here_backward += arr[i];
if(sum_till_here_backward > max_sum_backward){
max_sum_backward = sum_till_here_backward;
}
if(sum_till_here_backward < 0){
sum_till_here_backward=0;
}
sum_backward[i+1] = max_sum_backward;
}
for(int i=1; i<n; i++){
maximum=Math.max(maximum,(sum_forward[i-1]+sum_backward[i+1]));
}
}
System.out.print(maximum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
left=[]
for i in range(n):
left.append(0)
curr_max=l[0]
max_so_far=l[0]
for i in range(1,n):
left[i]=max_so_far
curr_max = max(l[i], curr_max+l[i])
max_so_far = max(max_so_far, curr_max)
right=[]
for i in range(n):
right.append(0)
curr_max=l[n-1]
max_so_far=l[n-1]
for i in range(n-2,-1,-1):
right[i]=max_so_far
curr_max = max(l[i], curr_max+l[i])
max_so_far = max(max_so_far, curr_max)
ans=0
for i in range(n):
ans=max(ans,left[i]+right[i])
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define ld long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
vector<int> a;
int n;
vector<int> kadane(){
int cur = 0;
int mx = 0;
vector<int> vect(n);
for(int i=0; i<n; i++){
cur += a[i];
mx = max(mx, cur);
if(cur < 0)
cur = 0;
vect[i]=mx;
}
return vect;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
cin>>n;
for(int i=0; i<n; i++){
int x; cin>>x;
a.pb(x);
}
vector<int> v1 = kadane();
reverse(all(a));
vector<int> v2 = kadane();
reverse(all(v2));
int ans = 0;
For(i, 0, n){
if(i>0 && i<n-1)
ans = max(ans, v1[i-1]+v2[i+1]);
else if(i==0)
ans = max(ans, v2[i+1]);
else
ans = max(ans, v1[i-1]);
}
cout<<ans;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a string S consisting of lowercase English letters. Determine whether adding some number of a's (possibly zero) at the beginning of S can make it a palindrome.The input consists of a single string S.
Constraints
1 ≤ ∣S∣ ≤ 10^6
S consists of lowercase English letters.If adding some number of a's (possibly zero) at the beginning of S can make it a palindrome, print Yes; otherwise, print No.<b>Sample Input 1</b>
kasaka
<b>Sample Output 1</b>
Yes
<b>Sample Input 2</b>
reveh
<b>Sample Output 2</b>
No, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(void) {
int n, x, y;
string a;
cin >> a;
n = a.size();
x = 0;
for (int i = 0; i < n; i++) {
if (a[i] == 'a')x++;
else break;
}
y = 0;
for (int i = n - 1; i >= 0; i--) {
if (a[i] == 'a')y++;
else break;
}
if (x == n) {
cout << "Yes" << endl;
return 0;
}
if (x > y) {
cout << "No" << endl;
return 0;
}
for (int i = x; i < (n - y); i++) {
if (a[i] != a[x + n - y - i - 1]) {
cout << "No" << endl;
return 0;
}
}
cout << "Yes" << endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: static void Icecreams (int N, int D){
int x=N;
while(D-->0){
x-=x/2;
x*=3;
}
System.out.println(x);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
cout << x;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: void Icecreams (int N, int D){
int x=N;
while(D--){
x-=x/2;
x*=3;
}
printf("%d", x);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is fond of ice cream initially she had N ice creams with her. If Sara eats exactly half of the ice cream she has in a day and the remaining icecreams get tripled each night. How many ice creams does Sara have at the end of D-days?
<b>Note:- </b>
Take the floor value while dividing by 2.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>Icecreams()</b> that takes integers N and D as parameters.
<b>Constraints:-</b>
1 <= N <= 100
1 <= D <= 20Print a single integer denoting the number of ice creams at the end of D days.Sample Input 1:-
5 1
Sample Output 1:-
9
</b>Explanation:-</b>
Sara will eat 2 ice creams and the remaining 3 will be tripled i. e 9.
Sample Input 2:-
5 3
Sample Output 2:-
24
<b>Explanation:-</b>
Day 1:- N=5, eaten = 2, rem = 3 => remaining = 3*3 = 9
Day 2:- N=9, eaten = 4, rem = 5 => remaining = 5*3 = 15
Day 3:- N=15, eaten = 7, rem = 8 => remaining = 8*3 = 24, I have written this Solution Code: def Icecreams(N,D):
ans = N
while D > 0:
ans = ans - ans//2
ans = ans*3
D = D-1
return ans
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main (String[] args) {
FastReader sc = new FastReader();
int n1 = sc.nextInt();
int n2 = sc.nextInt();
HashSet<Integer> hs = new HashSet<>();
long sum = 0l;
for(int i=0;i<n1;i++){
int curr = sc.nextInt();
hs.add(curr);
}
for(int i=0;i<n2;i++){
int sl = sc.nextInt();
if(hs.contains(sl)){
sum+=sl;
hs.remove(sl);
}
}
System.out.println(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: N1,N2=(map(int,input().split()))
arr1=set(map(int,input().split()))
arr2=set(map(int,input().split()))
arr1=list(arr1)
arr2=list(arr2)
arr1.sort()
arr2.sort()
i=0
j=0
sum1=0
while i<len(arr1) and j<len(arr2):
if arr1[i]==arr2[j]:
sum1+=arr1[i]
i+=1
j+=1
elif arr1[i]>arr2[j]:
j+=1
else:
i+=1
print(sum1), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Morgan was opening some boxes when she found one bearing a message “For Morgan, Dad loves you 3000". She wants to open the box but she needs to solve the question to open it. She asks Peter to do that. Help Peter and Morgan solve the question so that they can see what's inside the box.
Given two arrays Arr1 and Arr2 of size N1 and N2. Your task is to find the sum of all elements that are common to both arrays. If there are no common elements the output would be 0.
Note: The arrays may contain duplicate elements. However, you need to sum only unique elements that are common to both arrays.The first line of input contains N1 and N2 separated by a space. The second line contains N1 space separated elements of Arr1. The third line contains N2 space separated elements of Arr2.
Constraints:
1 <= N1, N2 <= 10<sup>6</sup>
1 <= Arr1[i], Arr2[i] <= 1000000000Print the sum of common elements.Sample Input:
5 6
1 2 3 4 5
2 3 4 5 6 7
Sample Output:
14
Explanation:- Common elements = 2, 3, 4 , 5
sum= 2 + 3 + 4 + 5 = 14
Sample Input:-
3 3
1 2 3
4 5 6
Sample Output:-
0, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define pu push_back
#define fi first
#define se second
#define mp make_pair
#define int long long
#define pii pair<int,int>
#define mm (s+e)/2
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define sz 200000
#define qw1 freopen("input1.txt", "r", stdin); freopen("output1.txt", "w", stdout);
#define qw2 freopen("input2.txt", "r", stdin); freopen("output2.txt", "w", stdout);
#define qw3 freopen("input3.txt", "r", stdin); freopen("output3.txt", "w", stdout);
#define qw4 freopen("input4.txt", "r", stdin); freopen("output4.txt", "w", stdout);
#define qw5 freopen("input5.txt", "r", stdin); freopen("output5.txt", "w", stdout);
#define qw6 freopen("input6.txt", "r", stdin); freopen("output6.txt", "w", stdout);
#define qw freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
signed main()
{
int n,m;
cin>>n>>m;
set<int> ss,yy;
for(int i=0;i<n;i++)
{
int a;
cin>>a;
ss.insert(a);
}
for(int i=0;i<m;i++)
{
int a;
cin>>a;
if(ss.find(a)!=ss.end()) yy.insert(a);
}
int sum=0;
for(auto it:yy)
{
sum+=it;
}
cout<<sum<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Implement <code>getObjKeys</code> which only takes one argument which will be a object.
The function should return all they keys present in object as a string where elements are seperated by a ', '. (No nested objects)(Use JS Built in function)Function will take one argument which will be an objectFunction will is string which contain all the keys from the input object seperated by a ', 'const obj = {email:"akshat. sethi@newtonschool. co", password:"123456"}
const keyString = getObjKeys(obj)
console. log(keyString) // prints email, password, I have written this Solution Code: function getObjKeys(obj){
return Object.keys(obj).join(",")
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
Stack <Integer> st = new Stack<>();
for(int i=0;i<t;i++){
char ch = str[i].charAt(0);
if(Character.isDigit(ch)){
st.push(Integer.parseInt(str[i]));
}
else{
int str1 = st.pop();
int str2 = st.pop();
switch(ch) {
case '+':
st.push(str2+str1);
break;
case '-':
st.push(str2- str1);
break;
case '/':
st.push(str2/str1);
break;
case '*':
st.push(str2*str1);
break;
}
}
}
System.out.println(st.peek());
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: stack = []
n = int(input())
exp = [i for i in input().split()]
for i in exp:
try:
stack.append(int(i))
except:
a1 = stack.pop()
a2 = stack.pop()
if i == '+':
stack.append(a1+a2)
if i == '-':
stack.append(a2-a1)
if i == '/':
stack.append(a2//a1)
if i == '*':
stack.append(a1*a2)
print(*stack), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a postfix expression, your task is to evaluate given expression.
Infix expression: The expression of the form a operator b. When an operator is in-between every pair of operands.
Postfix expression: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, -, *, /. Each operand may be an integer or another expression.The first line denotes the size of the string(which contains number or operand in form of string) i.e. N.
The next line contains string. The string contains an integer called operand or any of the four operators (+ - * /)
Constraints:-
1 <= n <= 40
1<=number<=500
Output the value of arithmetic expression formed using reverse Polish Notation.Input 1:
5
2 1 + 3 *
Output 1:
9
Explaination 1:
starting from backside:
*: ( )*( )
3: ()*(3)
+: ( () + () )*(3)
1: ( () + (1) )*(3)
2: ( (2) + (1) )*(3)
((2)+(1))*(3) = 9
Input 2:
5
4 13 5 / +
Output 2:
6
Explanation 2:
+: ()+()
/: ()+(() / ())
5: ()+(() / (5))
1: ()+((13) / (5))
4: (4)+((13) / (5))
(4)+((13) / (5)) = 6, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define int long long int
#define ld long double
#define pi pair<int, int>
#define pb push_back
#define fi first
#define se second
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0)
#ifndef LOCAL
#define endl '\n'
#endif
const int N = 2e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
class Solution {
public:
int evalRPN(vector<string> &tokens) {
int size = tokens.size();
if (size == 0)
return 0;
std::stack<int> operands;
for (int i = 0; i < size; ++i)
{
std::string cur_token = tokens[i];
if ((cur_token == "*") || (cur_token == "/") || (cur_token == "+") || (cur_token == "-"))
{
int opr2 = operands.top();
operands.pop();
int opr1 = operands.top();
operands.pop();
int result = this->eval(opr1, opr2, cur_token);
operands.push(result);
}
else{
operands.push(std::atoi(cur_token.c_str()));
}
}
return operands.top();
}
int eval(int opr1, int opr2, string opt)
{
if (opt == "*")
{
return opr1 * opr2;
}
else if (opt == "+")
{
return opr1 + opr2;
}
else if (opt == "-")
{
return opr1 - opr2;
}
else if (opt == "/")
{
return opr1 / opr2;
}
return 0;
}
};
signed main() {
IOS;
int n; cin >> n;
vector<string> v(n, "");
for(int i = 0; i < n; i++)
cin >> v[i];
Solution s;
cout << s.evalRPN(v);
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: static int numberOfDiagonal(int N){
if(N<=3){return 0;}
return (N*(N-3))/2;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code: def numberOfDiagonals(n):
if n <=3:
return 0
return (n*(n-3))//2
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. Write a program to the find number of diagonals possible in N sided convex polygon.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>numberOfDiagonal()</b> that takes the integer N parameter.
Constraints:
1 <= T <= 100
1 <= N <= 10^3Return number of diagonals possible in N sided convex polygon.Sample Input:
3
3
5
6
Sample Output:
0
5
9
Explanation:
For test case 2: The number of diagonals of 5 sided polygon: 5., I have written this Solution Code:
int numberOfDiagonals(int n){
if(n<=3){return 0;}
return (n*(n-3))/2;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
String str[]=br.readLine().split(" ");
long l=Long.parseLong(str[0]);
long r=Long.parseLong(str[1]);
long k=Long.parseLong(str[2]);
long count=(r/k)-((l-1)/k);
System.out.print(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: l,r,k=[int(x)for x in input().split()]
print(r//k - (l-1)//k), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You think that integers divisible by K are nice. Given L and R (L<=R), find the number of nice integers from L to R (both inclusive).First Line of input contains three integers L R K
Constraints :
0 <= L <= R <= 1000000000000000000(10^18)
1 <= K <= 1000000000000000000(10^18)Output a single integer which is the number of nice integers from L to R (both inclusive).Sample input 1
1 10 2
Sample output 1
5
Sample intput 2
4 5 3
Sample output 2
0
Explanation:-
Required divisors = 2 4 6 8 10, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
unsigned long long x,l,r,k;
cin>>l>>r>>k;
x=l/k;
if(l%k==0){x--;}
cout<<r/k-x;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String s = br.readLine();
if(s==null){
System.exit(0);
}
StringTokenizer st = new StringTokenizer(s, " ");
int power = Integer.parseInt(st.nextToken());
int multiple = Integer.parseInt(st.nextToken());
int res = power;
for(int i = 1;i<=multiple;i++){
res = res*2;
}
System.out.println(res);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: #include <iostream>
using namespace std;
int main()
{
int x, n;
cin >> x >> n;
cout << x*(1 << n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice's power is currently known to be an integer X. It is also known that her power doubles every second. For example, if Alice's power was currently 20, then after 2 seconds her power would have grown to 80.
Your task is to find out Alice's power after N seconds.The input consists of a single line containing two space-separated integers X and N.
<b>Constraints:</b>
1 ≤ X ≤ 1000
1 ≤ N ≤ 10Print a single integer – the power of Alice after N seconds.Sample Input 1:
5 2
Sample Output 1:
20
Sample Explanation 1:
Alice's power after 1 second will be 5*2 = 10. After 2 seconds it will be 10*2 = 20.
Sample Input 2:
4 3
Sample Output 2:
32, I have written this Solution Code: x,n = map(int,input().split())
print(x*(2**n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n=Integer.parseInt(br.readLine().trim());
if(n%3==1)
System.out.println("Dead");
else
System.out.println("Alive");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: n = int(input().strip())
if n%3 == 1:
print("Dead")
else:
print("Alive"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Joffrey has issued orders for Stark's execution. Now, you are trying to predict if Stark will survive or not. You are confident that Stark's survival depends on the N<sup>th</sup> fibonacci number's parity. If it is odd you will predict that Stark lives but if it is even you will predict that Stark dies. Given N, print your prediction.
Fibonacci series is a series where,
Fibonacci(1) = 0
Fibonacci(2) = 1
Fibonacci(i) = Fibonacci(i-1) + Fibonacci(i-2); for i > 2Input Contains a single integer N.
Constraints:
1 <= N <= 1000000Print "Alive" if Nth Fibonacci number is odd and print "Dead" if Nth Fibonacci number is even.Sample Input 1
3
Sample Output 1
Alive
Explanation: Fibonacci(3) = 1 which is odd.
Sample Input 2
4
Sample Output 1
Dead
Explanation: Fibonacci(4) = 2 which is even., I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
typedef unsigned long long ull;
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n%3==1)
cout<<"Dead";
else
cout<<"Alive";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: static int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: int SillyNumber(int N){
int sum=0;
int x=1;
while(sum<N){
sum+=x*x;
x++;
}
x--;
if(sum-N < N-(sum-x*x)){
return sum;
}
else{
return sum-x*x;
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A number is called Silly if it can be represented as the sum of the square of consecutive natural numbers starting from 1.
For a given number N, find the closest silly number.<b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>SillyNumber()</b> that takes integer N as argument.
Constraints:-
1 <= N <= 100000Return the closest Silly number.
Note:- If more than one answer exists return the minimum one.Sample Input:-
18
Sample Output:-
14
Explanation:-
1*1 + 2*2 + 3*3 = 14
Sample Input:-
2
Sample Output:-
1, I have written this Solution Code: def SillyNumber(N):
sum=0
x=1
while sum<N:
sum=sum+x*x
x=x+1
x=x-1
if (sum-N) < (N-(sum-x*x)):
return sum;
else:
return sum - x*x
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You have a box of paint. The paint can be used to cover area of 1 square unit. You have N magic spells with magic value from 1 to N. You can use each spell at max once.
When you use a spell on the box of paint, the quantity of paint in the box gets multiplied by the magic value of that spell. For example, if currently the paint in the box can cover A square units and a spell of B magic value is cast on the box, then after the spell the paint in the box can be used to cover A*B square units of area.
You want to paint a square (completely filled) of maximum area with integer side length with the paint such that after painting the square, the paint in the box gets empty. You can apply any magic spells you want in any order before the painting starts and not during or after the painting.
Find the area of maximum side square you can paint. As the answer can be huge find the area modulo 1000000007.The first and the only line of input contains a single integer N.
Constraints:
1 <= N <= 1000000Print the area of maximum side square you can paint modulo 1000000007.Sample Input 1
3
Sample Output 1
1
Explanation: We apply no magic spell.
Sample Input 2
4
Sample Output 2
4
Explanation: We apply magic spell number 4., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
int si[1000001];
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int ans=1;
int mo=1000000007;
for(int i=2;i<=n;++i){
if(si[i]==0){
int x=i;
int c=0;
while(x<=n){
c+=n/x;
x*=i;
}
if(c%2==0){
ans=(ans*i)%mo;
}
int j=i;
while(j<=n){
si[j]=1;
j+=i;
}
}
else{
ans=(ans*i)%mo;
}
}
cout<<ans;
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Yogi and Dharam are playing a game in which one has to solve the problem in order to get out of the room. Dharam goes first and Yogi gave him a problem. Since Dharam is having problem solving it, your task is to solve the problem for him.
Problem:-
Given a range from A to B your task is to count the numbers in which the difference between adjacent digits is not more than one.The first line of input contains a single integer T. The next T lines contain two space- separated integers each containing values of A and B.
Constraints:-
1 <= T <= 100
1 <= A, B <= 1000000000000For each test case print the count of numbers in the range [A, B] such that the difference between adjacent digits is not more than one.Sample Input:-
2
10 30
1 10
Sample Output:-
6
10
Explanation:-
10, 11, 12, 21, 22, 23
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static int count =0;
public static void solve(long a ,long b,long number){
if(number>b){
return;
}
if(number>=a && number <=b){
count++;
}
long last_digit =number % 10;
number *=10;
if(last_digit !=0){
solve(a,b,number+(last_digit-1));
}
if(last_digit !=9){
solve(a,b,number+(last_digit+1));
}
solve(a,b,number+(last_digit));
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0){
long a = sc.nextLong();
long b =sc.nextLong();
count =0;
for(long i=1l ; i<=9l ;i++){
solve(a,b,i);
}
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Yogi and Dharam are playing a game in which one has to solve the problem in order to get out of the room. Dharam goes first and Yogi gave him a problem. Since Dharam is having problem solving it, your task is to solve the problem for him.
Problem:-
Given a range from A to B your task is to count the numbers in which the difference between adjacent digits is not more than one.The first line of input contains a single integer T. The next T lines contain two space- separated integers each containing values of A and B.
Constraints:-
1 <= T <= 100
1 <= A, B <= 1000000000000For each test case print the count of numbers in the range [A, B] such that the difference between adjacent digits is not more than one.Sample Input:-
2
10 30
1 10
Sample Output:-
6
10
Explanation:-
10, 11, 12, 21, 22, 23
1, 2, 3, 4, 5, 6, 7, 8, 9, 10, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int cnt=0;
inline void solve(int a, int b, int p, int c, int i){
if(p>b){return;}
if(p>=a){cnt++;}
int x = p%10;
p*=10;
if(i==c){return;}
solve(a,b,p+x,c,i+1);
if(x!=9){
solve(a,b,p+x+1,c,i+1);
}
if(x!=0){
solve(a,b,p+x-1,c,i+1);
}
}
signed main(){
fast();
int t;
cin>>t;
while(t--){
int a,b;
cnt=0;
cin>>a>>b;
if(b<a){out(-1);continue;}
int ans=b;
int c=0;
while(ans){
c++;
ans/=10;
}
for(int i=1;i<=9;i++){
solve(a,b,i,c,1);
}
out(cnt);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void printMax(int arr[], int n, int k) {
int j, max;
for(int i = 0; i <= n - k; i++) {
max = arr[i];
for(j = 1; j < k; j++) {
if(arr[i + j] > max) {
max = arr[i + j];
}
}
System.out.print(max + " ");
}
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1[] = br.readLine().trim().split(" ");
int n = Integer.parseInt(str1[0]);
int k = Integer.parseInt(str1[1]);
String str2[] = br.readLine().trim().split(" ");
int arr[] = new int[n];
for(int i = 0; i < n; i++) {
arr[i] = Integer.parseInt(str2[i]);
}
printMax(arr, n ,k);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: n,k=input().split()
n=int(n)
k=int(k)
arr=input().split()
for i in range(0,n):
arr[i]=int(arr[i])
m=max(arr[0:k])
for i in range(k-1,n):
if(arr[i] > m):
m=arr[i]
if(arr[i-k]==m):
m=max(arr[i-k+1:i+1])
print (m, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A and an integer K. Find the maximum for each and every contiguous subarray of size K.
Problem asked in Amazon, Flipkart.The first line of each test case contains a single integer N denoting the size of array and the size of subarray K. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraints:
1 ≤ N ≤ 10^5
1 ≤ K ≤ N
0 ≤ A[i] <= 10^5Print the maximum for every subarray of size K.Sample Input:
9 3
1 2 3 1 4 5 2 3 6
Sample Output:
3 3 4 5 5 5 6
Explanation:
Starting from the first subarray of size k = 3, we have 3 as maximum. Moving the window forward, maximum element are as 3, 4, 5, 5, 5 and 6., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
// A Dequeue (Double ended queue) based method for printing maximum element of
// all subarrays of size k
void printKMax(int arr[], int n, int k)
{
// Create a Double Ended Queue, Qi that will store indexes of array elements
// The queue will store indexes of useful elements in every window and it will
// maintain decreasing order of values from front to rear in Qi, i.e.,
// arr[Qi.front[]] to arr[Qi.rear()] are sorted in decreasing order
std::deque<int> Qi(k);
/* Process first k (or first window) elements of array */
int i;
for (i = 0; i < k; ++i) {
// For every element, the previous smaller elements are useless so
// remove them from Qi
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back(); // Remove from rear
// Add new element at rear of queue
Qi.push_back(i);
}
// Process rest of the elements, i.e., from arr[k] to arr[n-1]
for (; i < n; ++i) {
// The element at the front of the queue is the largest element of
// previous window, so print it
cout << arr[Qi.front()] << " ";
// Remove the elements which are out of this window
while ((!Qi.empty()) && Qi.front() <= i - k)
Qi.pop_front(); // Remove from front of queue
// Remove all elements smaller than the currently
// being added element (remove useless elements)
while ((!Qi.empty()) && arr[i] >= arr[Qi.back()])
Qi.pop_back();
// Add current element at the rear of Qi
Qi.push_back(i);
}
// Print the maximum element of last window
cout << arr[Qi.front()];
}
// Driver program to test above functions
int main()
{
int n,k;
cin>>n>>k;
int arr[n];
for(int i=0;i<n;i++){
cin>>arr[i];
}
printKMax(arr, n, k);
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Modify the <code>takeMultipleNumbersAndAdd </code> such that it can take any number of arguments
and return its sum.
This is JS only question.Function should be able to take any number of argsSum of the numberstakeMultipleNumbersAndAdd(1, 2, 2) should return 5 because 1 + 2 + 2
takeMultipleNumbersAndAdd(-1, 2, -1, 5) should return 5, I have written this Solution Code:
function takeMultipleNumbersAndAdd (...nums){
// write your code here
return nums.reduce((prev,cur)=>prev+cur,0)
// return the output using return keyword
// do not console.log it
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of length N. Print the lexographically minimum rotation of the array Arr.
All the elements of the array are distinct.First line of input contains a single integer N.
Second line of input contains N integers denoting the array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000Print the lexographically minimum rotation of the array Arr.Sample Input
5
2 3 1 4 10
Sample Output
1 4 10 2 3, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
br.readLine();
String[] line = br.readLine().split(" ");
int minIndex = 0;
long minVal = Long.MAX_VALUE;
for (int i=0;i<line.length;++i){
long el = Long.parseLong(line[i]);
if (minVal>el){
minVal = el;
minIndex = i;
}
}
StringBuilder sb = new StringBuilder();
for (int i = minIndex;i< line.length;++i){
sb.append(line[i]+" ");
}
for (int i=0;i<minIndex;++i){
sb.append(line[i]+" ");
}
System.out.print(sb);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of length N. Print the lexographically minimum rotation of the array Arr.
All the elements of the array are distinct.First line of input contains a single integer N.
Second line of input contains N integers denoting the array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000Print the lexographically minimum rotation of the array Arr.Sample Input
5
2 3 1 4 10
Sample Output
1 4 10 2 3, I have written this Solution Code: N = int(input())
arr = list(map(int, input().split()))
mi = arr.index(min(arr))
ans = arr[mi:] + arr[:mi]
for e in ans:
print(e, end=' '), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array Arr of length N. Print the lexographically minimum rotation of the array Arr.
All the elements of the array are distinct.First line of input contains a single integer N.
Second line of input contains N integers denoting the array Arr.
Constraints:
1 <= N <= 100000
1 <= Arr[i] <= 1000000000Print the lexographically minimum rotation of the array Arr.Sample Input
5
2 3 1 4 10
Sample Output
1 4 10 2 3, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
int a[n];
for(int i=0;i<n;++i){
cin>>a[i];
}
int mi=0;
for(int i=0;i<n;++i)
if(a[i]<a[mi])
mi=i;
for(int i=0;i<n;++i)
cout<<a[(i+mi)%n]<<" ";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice like grids. Today, she is playing a computer game on an N*M grid.
The rows of the grid are numbered from 1 to N and the columns are numbered from 1 to M. The cell in row A and column B is represented as (A, B).
The grid has K lasers in it. Each laser is fixed at a single cell of the grid and can either defend the whole row it is placed in, or the whole column it is placed in.
Alice's player starts on the cell (1, 1) at time t = 0. Every second the following process takes place:
<ul>
<li> Alice chooses a neighbouring cell (a cell that shares an edge with the current cell) and moves her player to that cell. She cannot choose to remain on the same cell. </li>
<li> All the lasers flip their orientation i. e. if a laser was defending its row, then it now defends its column and vice versa. </li>
<li> If the cell the player landed on is defended by any laser, the player dies and Alice loses instantly. </li>
</ul>
Note that the player dies only if the cell is defended in <b>the new orientation of the lasers</b>.
The player is allowed to visit the same cell any number of times.
Determine the minimum time it will take Alice to navigate the grid and reach cell (N, M) without losing.
It is guaranteed that no lasers are placed on rows 1 and N, and columns 1 and M. However, a laser may defend cells on these rows and columns.The first line of input contains three integers: N, M and K. (3 <= N <= 3000, 3 <= M <= 3000, 1 <= K <= 3000)
The next K lines each contain the description of a laser's initial state (at time t = 0). The ith line contains two integers, x<sub>i</sub> and y<sub>i</sub>; followed by a character, O<sub>i</sub>.
2 <= x<sub>i</sub> <= N-1
2 <= y<sub>i</sub> <= M-1
The ith laser is stationed on the cell (x<sub>i</sub>, y<sub>i</sub>) and its initial orientation is O<sub>i</sub>. O<sub>i</sub> is either 'R' or 'C' denoting whether the laser defends its row or column respectively.Print a single integer, the minimum possible time for Alice to make her player reach (N, M). If it is impossible to reach (N, M) print -1.Sample Input 1:
4 4 1
2 3 R
Sample Output 1:
6
Sample Input 2:
3 3 1
2 2 R
Sample Output 2:
-1
Explanation for Sample 2:
At time t = 0, the laser defends (2, 1), (2, 2) and (2, 3).
At time t = 1, the laser defends (1, 2), (2, 2) and (3, 2).
and so on..
It can be easily verified that the player is always forced to move in the first column and hence, can never reach (3, 3).
, I have written this Solution Code: //HEADER FILES AND NAMESPACES
#include<bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// #include <sys/resource.h>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
template <typename T>
using ordered_multiset = tree<T, null_type, less_equal<T>, rb_tree_tag, tree_order_statistics_node_update>;
// DEFINE STATEMENTS
const long long infty = 1e18;
#define num1 1000000007
#define num2 998244353
#define REP(i,a,n) for(ll i=a;i<n;i++)
#define REPd(i,a,n) for(ll i=a; i>=n; i--)
#define pb push_back
#define pob pop_back
#define f first
#define s second
#define fix(f,n) std::fixed<<std::setprecision(n)<<f
#define all(x) x.begin(), x.end()
#define M_PI 3.14159265358979323846
#define epsilon (double)(0.000000001)
#define popcount __builtin_popcountll
#define fileio(x) freopen("input.txt", "r", stdin); freopen(x, "w", stdout);
#define out(x) cout << ((x) ? "Yes\n" : "No\n")
#define sz(x) x.size()
#define start_clock() auto start_time = std::chrono::high_resolution_clock::now();
#define measure() auto end_time = std::chrono::high_resolution_clock::now(); cerr << (end_time - start_time)/std::chrono::milliseconds(1) << "ms" << endl;
typedef long long ll;
typedef vector<long long> vll;
typedef pair<long long, long long> pll;
typedef vector<pair<long long, long long>> vpll;
typedef vector<int> vii;
// DEBUG FUNCTIONS
#ifdef LOCALY
template<typename T>
void __p(T a) {
cout<<a;
}
template<typename T, typename F>
void __p(pair<T, F> a) {
cout<<"{";
__p(a.first);
cout<<",";
__p(a.second);
cout<<"}";
}
template<typename T>
void __p(std::vector<T> a) {
cout<<"{";
for(auto it=a.begin(); it<a.end(); it++)
__p(*it),cout<<",}"[it+1==a.end()];
}
template<typename T>
void __p(std::set<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T>
void __p(std::multiset<T> a) {
cout<<"{";
for(auto it=a.begin(); it!=a.end();){
__p(*it);
cout<<",}"[++it==a.end()];
}
}
template<typename T, typename F>
void __p(std::map<T,F> a) {
cout<<"{\n";
for(auto it=a.begin(); it!=a.end();++it)
{
__p(it->first);
cout << ": ";
__p(it->second);
cout<<"\n";
}
cout << "}\n";
}
template<typename T, typename ...Arg>
void __p(T a1, Arg ...a) {
__p(a1);
__p(a...);
}
template<typename Arg1>
void __f(const char *name, Arg1 &&arg1) {
cout<<name<<" : ";
__p(arg1);
cout<<endl;
}
template<typename Arg1, typename ... Args>
void __f(const char *names, Arg1 &&arg1, Args &&... args) {
int bracket=0,i=0;
for(;; i++)
if(names[i]==','&&bracket==0)
break;
else if(names[i]=='(')
bracket++;
else if(names[i]==')')
bracket--;
const char *comma=names+i;
cout.write(names,comma-names)<<" : ";
__p(arg1);
cout<<" | ";
__f(comma+1,args...);
}
#define trace(...) cout<<"Line:"<<__LINE__<<" ", __f(#__VA_ARGS__, __VA_ARGS__)
#else
#define trace(...)
#define error(...)
#endif
// DEBUG FUNCTIONS END
// CUSTOM HASH TO SPEED UP UNORDERED MAP AND TO AVOID FORCED CLASHES
struct custom_hash {
static uint64_t splitmix64(uint64_t x) {
x += 0x9e3779b97f4a7c15;
x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
return x ^ (x >> 31);
}
size_t operator()(uint64_t x) const {
static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
return splitmix64(x + FIXED_RANDOM);
}
};
mt19937_64 rng(chrono::steady_clock::now().time_since_epoch().count()); // FOR RANDOM NUMBER GENERATION
ll mod_exp(ll a, ll b, ll c)
{
ll res=1; a=a%c;
while(b>0)
{
if(b%2==1)
res=(res*a)%c;
b/=2;
a=(a*a)%c;
}
return res;
}
ll mymod(ll a,ll b)
{
return (((a = a%b) < 0) ? a + b : a);
}
ll gcdExtended(ll,ll,ll *,ll *);
ll modInverse(ll a, ll m)
{
ll x, y;
ll g = gcdExtended(a, m, &x, &y);
g++; //this line was added just to remove compiler warning
ll res = (x%m + m) % m;
return res;
}
ll gcdExtended(ll a, ll b, ll *x, ll *y)
{
if (a == 0)
{
*x = 0, *y = 1;
return b;
}
ll x1, y1;
ll gcd = gcdExtended(b%a, a, &x1, &y1);
*x = y1 - (b/a) * x1;
*y = x1;
return gcd;
}
struct Graph
{
vector<vector<int>> adj;
Graph(int n)
{
adj.resize(n+1);
}
void add_edge(int a, int b, bool directed = false)
{
adj[a].pb(b);
if(!directed) adj[b].pb(a);
}
};
ll bfs(vector<vector<bool>> &allowed) {
ll n = allowed.size();
ll m = allowed[0].size();
queue<pll> q;
vector<vector<int>> dist(n, vector<int>(m, -1));
q.push({0, 0});
dist[0][0] = 0;
ll dx[] = {1, -1, 0, 0};
ll dy[] = {0, 0, 1, -1};
while(!q.empty()) {
pll f = q.front();
q.pop();
REP(i, 0, 4)
{
ll nx = f.first + dx[i];
ll ny = f.second + dy[i];
if(0 <= nx && nx < n && 0 <= ny && ny < m && dist[nx][ny] == -1 && allowed[nx][ny]) {
q.push({nx, ny});
dist[nx][ny] = dist[f.first][f.second] + 1;
}
}
}
return dist[n-1][m-1];
}
void show_grid(vector<vector<bool>> allowed) {
for(auto &x: allowed)
{
for(auto y: x)
{
cout << y;
}
cout << "\n";
}
}
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
// freopen("test-4.txt", "r", stdin);
// fileio("output.txt");
ll n, m, k;
cin >> n >> m >> k;
vector<bool> row_odd(n, false), row_even(n, false), col_odd(m, false), col_even(m, false);
vector<vector<bool>> allowed(n, vector<bool>(m, true));
REP(i, 0, k)
{
ll a, b;
char c;
cin >> a >> b >> c;
a--; b--;
if(c == 'R') {
col_odd[b] = true;
row_even[a] = true;
}
else {
col_even[b] = true;
row_odd[a] = true;
}
allowed[a][b] = false;
}
REP(i, 0, n)
{
REP(j, 0, m)
{
if((i+j)%2 == 0 && (row_even[i] || col_even[j])) {
allowed[i][j] = false;
}
if((i+j)%2 == 1 && (row_odd[i] || col_odd[j])) {
allowed[i][j] = false;
}
}
}
// show_grid(allowed);
cout << bfs(allowed) << "\n";
return 0;
}
/*
1. Check borderline constraints. Can a variable you are dividing by be 0?
2. Use ll while using bitshifts
3. Do not erase from set while iterating it
4. Initialise everything
5. Read the task carefully, is something unique, sorted, adjacent, guaranteed??
6. DO NOT use if(!mp[x]) if you want to iterate the map later
7. Are you using i in all loops? Are the i's conflicting?
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a linked list consisting of N nodes, your task is to delete every kth Node from the circular linked list until only one node is left. Also, print the intermediate lists
<b>Note:
Sample Input and Output just show how a linked list will look depending on the questions. Do not copy-paste as it is in custom input</b><b>User Task:</b>
Since this will be a functional problem, you don't have to take input. You just have to complete the function <b>deleteK()</b> that takes head node of circular linked list and the integer K as parameter.
Constraints:
1 <= K <= N <= 500
1 <= Node. data<= 1000Print the intermediate nodes until one node is left as shown in example.Sample Input:-
4 2
1 2 3 4
Sample Output:-
1->2->3->4->1
1->2->4->1
2->4->2
2->2
Sample Input:-
9 4
1 2 3 4 5 6 7 8 9
Sample Output:-
1->2->3->4->5->6->7->8->9->1
1->2->3->4->6->7->8->9->1
1->2->3->4->6->7->8->1
1->2->3->6->7->8->1
2->3->6->7->8->2
2->3->6->8->2
2->3->8->2
2->3->2
2->2, I have written this Solution Code: static void printList(Node head)
{
if (head == null)
return;
Node temp = head;
do
{
System.out.print( temp.data + "->");
temp = temp.next;
}
while (temp != head);
System.out.println(head.data );
}
/*Function to delete every kth Node*/
static Node deleteK(Node head_ref, int k)
{
Node head = head_ref;
// If list is empty, simply return.
if (head == null)
return null;
// take two pointers - current and previous
Node curr = head, prev=null;
while (true)
{
// Check if Node is the only Node\
// If yes, we reached the goal, therefore
// return.
if (curr.next == head && curr == head)
break;
// Print intermediate list.
printList(head);
// If more than one Node present in the list,
// Make previous pointer point to current
// Iterate current pointer k times,
// i.e. current Node is to be deleted.
for (int i = 0; i < k; i++)
{
prev = curr;
curr = curr.next;
}
// If Node to be deleted is head
if (curr == head)
{
prev = head;
while (prev.next != head)
prev = prev.next;
head = curr.next;
prev.next = head;
head_ref = head;
}
// If Node to be deleted is last Node.
else if (curr.next == head)
{
prev.next = head;
}
else
{
prev.next = curr.next;
}
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N line segments on the 2-D plane. The endpoints of the i<sup>th</sup> line segment are (0, i) and (1, P<sub>i</sub>). Now your task is to highlight all the line segments so that small Tono can analyze them properly. The cost of highlighting the i<sup>th</sup> line segment is C<sub>i</sub>.
Now there is a catch, if you choose to highlight the k<sup>th</sup> line segment, <b>ALL</b> the line segments intersecting the k<sup>th</sup> line segment are also highlighted (there is no further propogation). You <b>CANNOT</b> choose a line segment that has already been highlighted.
Find the minimum cost of highlighting all the line segments.
Note: P is the permutation of the first N natural numbers.The first line of the input contains an integer N.
The second line of the input contains N integers, P<sub>1</sub>, P<sub>2</sub>,... , P<sub>N</sub>.
The third line of the input contains N integers, C<sub>1</sub>, C<sub>2</sub>,... , C<sub>N</sub>.
Constraints
1 <= N <= 100000
1 <= P<sub>i</sub> <= N (P is a permutation of first N natural numbers)
1 <= C<sub>i</sub> <= 20000Output a single integer, the minimum cost to highlight all the line segments.Sample Input
4
1 2 3 4
1 2 3 4
Sample Output
10
Explanation: You need to highlight all the line segments individually.
Sample Input
4
2 1 4 3
1 2 3 4
Sample Output
4, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(int i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define ld long double
#define int long long
#define double long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
const int MOD = 1e9+7;
const int INF = 1LL<<60;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
const int N = 1e5 + 5;
const int SQN = 1 << 10;
const int SZ = (N / SQN) + 5;
int n;
int p[N];
int w[N];
int r[N];
int dp[N];
int dp2[N];
int stk[SZ][SQN];
int sz[SZ];
void update(int row , int col , int val){
int where = col >> 10;
while(sz[where] && p[stk[where][sz[where]]] < col){
--sz[where];
}
stk[where][++sz[where]] = row;
dp2[row] = val;
for(int i = sz[where] - 1 ; i >= 1 ; --i){
val = min(val , dp2[stk[where][i]] = min(dp[stk[where][i]] , val));
}
}
int query(int row , int col){
int mx = -1;
int res = 2e9 + 9;
for(int i = col - 1 ; i >= ((col >> 10) << 10) ; --i){
int ro = r[i];
if(ro < row){
if(ro > mx){
res = min(res , dp[ro]);
mx = ro;
}
}
}
for(int block = (col >> 10) - 1 ; block >= 0 ; --block){
if(sz[block] && stk[block][sz[block]] > mx){
int l = 1;
int r = sz[block];
while(l < r){
int mid = (l + r) >> 1;
if(stk[block][mid] > mx){
r = mid;
}
else{
l = mid + 1;
}
}
res = min(res , dp2[stk[block][l]]);
mx = stk[block][sz[block]];
}
}
return res;
}
signed main(){
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
cin>>n;
for(int i = 1 ; i <= n ; ++i){
cin>>p[i];
r[p[i]] = i;
}
for(int i = 1 ; i <= n ; ++i){
cin>>w[i];
}
p[0] = 0;
r[0] = 0;
p[n + 1] = n + 1;
r[n + 1] = n + 1;
w[0] = 0;
w[n + 1] = 0;
dp[0] = 0;
update(0 , 0 , 0);
for(int i = 1 ; i <= n + 1 ; ++i){
dp[i] = query(i , p[i]) + w[i];
update(i , p[i] , dp[i]);
}
cout<<dp[n+1];
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code:
import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String str1 = br.readLine();
String str2 = br.readLine();
countCommonCharacters(str1, str2);
}
static void countCommonCharacters(String s1, String s2) {
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for (int i = 0; i < s1.length(); i++)
arr1[s1.codePointAt(i) - 97]++;
for (int i = 0; i < s2.length(); i++)
arr2[s2.codePointAt(i) - 97]++;
int lenToCut = 0;
for (int i = 0; i < 26; i++) {
int leastOccurrence = Math.min(arr1[i], arr2[i]);
lenToCut += (2 * leastOccurrence);
}
System.out.println(findRelation((s1.length() + s2.length() - lenToCut) % 6));
}
static String findRelation(int value) {
switch (value) {
case 1:
return "Friends";
case 2:
return "Love";
case 3:
return "Affection";
case 4:
return "Marriage";
case 5:
return "Enemy";
default:
return "Siblings";
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: name1 = input().strip().lower()
name2 = input().strip().lower()
listA = [0]*26
listB = [0]*26
for i in name1:
listA[ord(i)-ord('a')] = listA[ord(i)-ord('a')] + 1
for i in name2:
listB[ord(i)-ord('a')] = listB[ord(i)-ord('a')] + 1
count = 0
for i in range(0,26):
count = count+abs(listA[i]-listB[i])
res = ["Siblings","Friends","Love","Affection", "Marriage", "Enemy"]
print(res[count%6]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: They say friendship is greater than love. Why not play the famous game "FLAMES".
The rules are super simple. Given two strings (all lowercase), remove all the letters that are common to both the strings from both the strings. You cannot erase a character in first string whichever corresponding same character in other string not exist.
For example, in the case
String 1: saumya
String 2: ansh
You can remove only 1 'a' and 1 's' from both the strings.
Remaining strings are:
String 1: umya
String 2: nh
Now all you need to do is find the sum of the remaining strings length % 6.
Output:
If obtained value is 1, output "Friends"
If obtained value is 2, output "Love"
If obtained value is 3, output "Affection"
If obtained value is 4, output "Marriage"
If obtained value is 5, output "Enemy"
If obtained value is 0, output "Siblings"You will be given two strings on different lines.
Constraints
1 <= Length of both the strings <= 100000Output a single string, the result of FLAMES test.Sample Input:-
saumya
ansh
Sample Output:-
Siblings
Explanation:-
after deleting characters :-
str1 = umya
str2 = nh
sum = 4+2
sum%6=0
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
string s1,s2;
cin>>s1>>s2;
string a[6];
a[1]= "Friends";
a[2]= "Love";
a[3]="Affection";
a[4]= "Marriage";
a[5]= "Enemy";
a[0]= "Siblings";
int b[26],c[26];
for(int i=0;i<26;i++){b[i]=0;c[i]=0;}
for(int i=0;i<s1.length();i++){
b[s1[i]-'a']++;
}
for(int i=0;i<s2.length();i++){
c[s2[i]-'a']++;
}
int sum=0;
for(int i=0;i<26;i++){
sum+=abs(b[i]-c[i]);
}
cout<<a[sum%6];
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(r);
String a = in.readLine();
String[] nums = a.split(" ");
long[] l = new long[3];
for(int i=0; i<3; i++){
l[i] = Long.parseLong(nums[i]);
}
Arrays.sort(l);
System.out.print(l[1]);
}
catch(Exception e){
System.out.println(e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code:
//#define ASC
//#define DBG_LOCAL
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#define int long long
// #define int __int128
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename T>
using v = vector<T>;
template <typename T>
using vv = vector<vector<T>>;
template <typename T>
using vvv = vector<vector<vector<T>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1LL * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double PI = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % 2;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
void solv()
{
int A ,B, C;
cin>>A>>B>>C;
vector<int> values;
values.push_back(A);
values.push_back(B);
values.push_back(C);
sort(all(values));
cout<<values[1]<<endl;
}
void solve()
{
int t = 1;
// cin>>t;
for(int i = 1;i<=t;i++)
{
// cout<<"Case #"<<i<<": ";
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#else
#ifdef ASC
namespace fs = std::filesystem;
std::string path = "./";
string filename;
for (const auto & entry : fs::directory_iterator(path)){
if( entry.path().extension().string() == ".in"){
filename = entry.path().filename().stem().string();
}
}
if(filename != ""){
string input_file = filename +".in";
string output_file = filename +".out";
if (fopen(input_file.c_str(), "r"))
{
freopen(input_file.c_str(), "r", stdin);
freopen(output_file.c_str(), "w", stdout);
}
}
#endif
#endif
// auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
// cout<<endl;
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
// clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: lst = list(map(int, input().split()))
lst.sort()
print(lst[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays of size N which have the same values but in different orders, we need to make a second array the same as a first array using a minimum number of swaps.
Note:- It is guaranteed that the elements of the array are uniqueFirst line of input contains the size of array N, second line of input contains N space separated integers depicting values of first array, third line of input contains N space separated integers depicting values of second array.
Constraints:-
1 < = N < = 10000
1 < = Arr[i] < = 1000000000Print the minimum number of swaps required to make the second array equal to first.Sample Input:-
5
1 2 3 4 5
3 1 2 5 4
Sample Output:-
3
Sample Input:-
4
3 6 4 8
4 6 8 3
Sample Output:-
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int n = Integer.parseInt(br.readLine());
String s[] = br.readLine().split(" ");
int a[] = new int[n];
int b[] = new int[n];
for(int i=0;i<n;i++)
a[i] = Integer.parseInt(s[i]);
String s1[] = br.readLine().split(" ");
for(int i=0;i<n;i++)
b[i] = Integer.parseInt(s1[i]);
int t = miniSwap(a,b,n);
System.out.print(t);
}
public static int miniSwap(int []a, int b[], int n){
int count=0;
for(int i=0;i<n;i++){
if(a[i]!=b[i]){
count++;
for(int j=i+1;j<n;j++){
if(a[i]==b[j])
swap(i,j,b);
}
}
}
return count;
}
public static void swap(int i,int j,int []b){
int temp = b[i];
b[i] = b[j];
b[j] = temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two arrays of size N which have the same values but in different orders, we need to make a second array the same as a first array using a minimum number of swaps.
Note:- It is guaranteed that the elements of the array are uniqueFirst line of input contains the size of array N, second line of input contains N space separated integers depicting values of first array, third line of input contains N space separated integers depicting values of second array.
Constraints:-
1 < = N < = 10000
1 < = Arr[i] < = 1000000000Print the minimum number of swaps required to make the second array equal to first.Sample Input:-
5
1 2 3 4 5
3 1 2 5 4
Sample Output:-
3
Sample Input:-
4
3 6 4 8
4 6 8 3
Sample Output:-
2, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define int long long int
// Function returns the minimum number of swaps
// required to sort the array
// This method is taken from below post
// https://www.geeksforgeeks.org/minimum-number-swaps-required-sort-array/
int minSwapsToSort(int arr[], int n)
{
// Create an array of pairs where first
// element is array element and second element
// is position of first element
pair<int, int> arrPos[n];
for (int i = 0; i < n; i++)
{
arrPos[i].first = arr[i];
arrPos[i].second = i;
}
// Sort the array by array element values to
// get right position of every element as second
// element of pair.
sort(arrPos, arrPos + n);
// To keep track of visited elements. Initialize
// all elements as not visited or false.
vector<bool> vis(n, false);
// Initialize result
int ans = 0;
// Traverse array elements
for (int i = 0; i < n; i++)
{
// already swapped and corrected or
// already present at correct pos
if (vis[i] || arrPos[i].second == i)
continue;
// find out the number of node in
// this cycle and add in ans
int cycle_size = 0;
int j = i;
while (!vis[j])
{
vis[j] = 1;
// move to next node
j = arrPos[j].second;
cycle_size++;
}
// Update answer by adding current cycle.
ans += (cycle_size - 1);
}
// Return result
return ans;
}
// method returns minimum number of swap to make
// array B same as array A
int minSwapToMakeArraySame(int a[], int b[], int n)
{
// map to store position of elements in array B
// we basically store element to index mapping.
map<int, int> mp;
for (int i = 0; i < n; i++)
mp[b[i]] = i;
// now we're storing position of array A elements
// in array B.
for (int i = 0; i < n; i++)
b[i] = mp[a[i]];
/* We can uncomment this section to print modified
b array
for (int i = 0; i < N; i++)
cout << b[i] << " ";
cout << endl; */
// returing minimum swap for sorting in modified
// array B as final answer
return minSwapsToSort(b, n);
}
// Driver code to test above methods
signed main()
{
int n;
cin>>n;
int a[n];
int b[n];
for(int i=0;i<n;i++){
cin>>a[i];}
for(int i=0;i<n;i++){
cin>>b[i];
}
cout << minSwapToMakeArraySame(a, b, n);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<b>User task:</b> Since this is a functional problem you don't have to worry about the input. You just have to complete the function kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: void kCircleSum(int arr[],int n,int k){
long long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
printf("%lli ",ans);
ans+=arr[(i+k)%n]-arr[i];
}
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<b>User task:</b> Since this is a functional problem you don't have to worry about the input. You just have to complete the function kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: function kCircleSum(arr, arrSize, k)
{
var list = new Array(2*arrSize + 5)
for(var i = 0; i < arrSize; i++)
{
list[i+1] = arr[i]
list[i+arrSize+1] = list[i+1]
}
for(var i = 0; i < 2*arrSize; i++)
dp[i] = 0
for(var i=1;i<=2*arrSize;i++)
{
dp[i] = dp[i-1]+list[i]
}
var ans = ""
for(var i = 1; i <= arrSize; i++)
{
ans += (dp[i+k-1]-dp[i-1]) + " "
}
console.log(ans)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<b>User task:</b> Since this is a functional problem you don't have to worry about the input. You just have to complete the function kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: static void kCircleSum(int arr[],int n,int k){
long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
System.out.print(ans+" ");
ans+=arr[(i+k)%n]-arr[i];
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<b>User task:</b> Since this is a functional problem you don't have to worry about the input. You just have to complete the function kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: def kCircleSum(arr,n,k):
ans=0
for i in range (0,k):
ans=ans+arr[i]
for i in range (0,n):
print(ans,end=" ")
ans=ans+arr[int((i+k)%n)]-arr[i]
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N numbers are arranged in Circle. Find the Sum of all K contiguous Sub-arrays.The first line of the input contains an integer N, the length of the array, and K. The next line contains N integers which are elements of the array.
<b>User task:</b> Since this is a functional problem you don't have to worry about the input. You just have to complete the function kCircleSum(arr, N, K) which contains arr(array) and N(size of the array), and K as a parameter
<b>Constraints</b>
1 <= N <= 100000
1 <= arr[I] <= 100000
1 <= K <= NYou need to print N space-separated integers ith integer denoting Sum of sub-array of length K starting at index i.Sample Input
3 1
1 2 3
Sample Output
1 2 3
Explanation : k=1 so ans is 1, 2, and 3.
Sample Input
5 2
6 4 3 4 1
Sample Output
10 7 7 5 7, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
void kCircleSum(int arr[],int n,int k){
long long ans=0;
for(int i=0;i<k;i++){
ans+=arr[i];
}
for(int i=0;i<n;i++){
printf("%lli ",ans);
ans+=arr[(i+k)%n]-arr[i];
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Priyanka works for an international toy company that ships by container. Her task is to the determine the lowest cost way to combine her orders for shipping. She has a list of item weights. The shipping company has a requirement that all items loaded in a container must weigh less than or equal to 4 units plus the weight of the minimum weight item. All items meeting that requirement will be shipped in one container.
What is the smallest number of containers that can be contracted to ship the items based on the given list of weights?
See example for better understanding.The first line contains an integer N, the number of orders to ship.
The next line contains N space- separated integers, w[1], w[2]., w[N] representing the orders in a weight array.
1 <= N <= 100000
1 <= w[i] <= 100000Print the integer value of the number of containers Priyanka must contract to ship all of the toys.Sample Input
9
1 2 3 4 5 10 11 12 13
Sample Output
2
There are items with weights w = [1, 2, 3, 4, 5, 10, 11, 12, 13]. This can be broken into two containers: [1, 2, 3, 4, 5] and [10, 11, 12, 13]. Each container will contain items weighing within 4 units of the minimum weight item., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
private static Reader in;
private static PrintWriter out;
public static void main(String[] args) throws IOException {
in = new Reader();
out = new PrintWriter(System.out, true);
int N = in.nextInt();
int[] arr = new int[N];
for (int i = 0; i < N; i++)arr[i] = in.nextInt();
Arrays.sort(arr);
int last = -10000;
int count = 0;
for (int i = 0; i < N; i++) {
if (last+4 < arr[i]) {
last = arr[i];
count++;
}
}
out.println(count);
out.close();
System.exit(0);
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException {
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
byte[] buf = new byte[1024];
int cnt = 0;
byte c = read();
while (c <= ' ')
c = read();
do {
buf[cnt++] = c;
} while ((c = read()) != '\n');
return new String(buf, 0, cnt);
}
public String next() throws IOException {
byte[] buf = new byte[1024];
int cnt = 0;
byte c = read();
while (c <= ' ')
c = read();
do {
buf[cnt++] = c;
} while ((c = read()) > ' ');
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.')
while ((c = read()) >= '0' && c <= '9')
ret += (c - '0') / (div *= 10);
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException {
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException {
if (din == null)
return;
din.close();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Priyanka works for an international toy company that ships by container. Her task is to the determine the lowest cost way to combine her orders for shipping. She has a list of item weights. The shipping company has a requirement that all items loaded in a container must weigh less than or equal to 4 units plus the weight of the minimum weight item. All items meeting that requirement will be shipped in one container.
What is the smallest number of containers that can be contracted to ship the items based on the given list of weights?
See example for better understanding.The first line contains an integer N, the number of orders to ship.
The next line contains N space- separated integers, w[1], w[2]., w[N] representing the orders in a weight array.
1 <= N <= 100000
1 <= w[i] <= 100000Print the integer value of the number of containers Priyanka must contract to ship all of the toys.Sample Input
9
1 2 3 4 5 10 11 12 13
Sample Output
2
There are items with weights w = [1, 2, 3, 4, 5, 10, 11, 12, 13]. This can be broken into two containers: [1, 2, 3, 4, 5] and [10, 11, 12, 13]. Each container will contain items weighing within 4 units of the minimum weight item., I have written this Solution Code: import sys
N=int(sys.stdin.readline())
a=list(sys.stdin.readline().split())
for index,item in enumerate(a):
a[index]=int(item)
a=sorted(a)
count=0
i=0
while i<N:
temp=int(a[i])+4
count+=1
while i<N and int(a[i])<=temp:
i+=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Priyanka works for an international toy company that ships by container. Her task is to the determine the lowest cost way to combine her orders for shipping. She has a list of item weights. The shipping company has a requirement that all items loaded in a container must weigh less than or equal to 4 units plus the weight of the minimum weight item. All items meeting that requirement will be shipped in one container.
What is the smallest number of containers that can be contracted to ship the items based on the given list of weights?
See example for better understanding.The first line contains an integer N, the number of orders to ship.
The next line contains N space- separated integers, w[1], w[2]., w[N] representing the orders in a weight array.
1 <= N <= 100000
1 <= w[i] <= 100000Print the integer value of the number of containers Priyanka must contract to ship all of the toys.Sample Input
9
1 2 3 4 5 10 11 12 13
Sample Output
2
There are items with weights w = [1, 2, 3, 4, 5, 10, 11, 12, 13]. This can be broken into two containers: [1, 2, 3, 4, 5] and [10, 11, 12, 13]. Each container will contain items weighing within 4 units of the minimum weight item., I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ll long long
#define pb push_back
#define endl '\n'
#define pii pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define F first
#define S second
#define sz(x) (int)x.size()
#define hell 1000000007
#define rep(i,a,b) for(int i=a;i<b;i++)
#define dep(i,a,b) for(int i=a;i>=b;i--)
#define lbnd lower_bound
#define ubnd upper_bound
#define bs binary_search
#define mp make_pair
using namespace std;
const int N = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
void solve(){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
int ans = 0, l = 1;
a[n+1] = inf;
for(int i = 1; i <= n+1; i++){
if(a[i]-a[l] > 4)
ans++, l = i;
}
cout << ans;
}
void testcases(){
int tt = 1;
//cin >> tt;
while(tt--){
solve();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
clock_t start = clock();
testcases();
cerr << (double)(clock() - start)*1000/CLOCKS_PER_SEC << " ms: ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Aarav is standing at location X and Reena is standing at location Y. In one step Aarav can increase or decrease his location by at most A. What is the minimum no. of steps required by Aarav to reach Reena?The first line of input will contain a single integer T, denoting the number of test cases. Each test case consists of three integers X, Y, and A, the initial coordinate of Aarav, the initial coordinate of Reena and the maximum number of coordinates Aarav can move in one step.
<b>Constraints</b>
1 ≤ T ≤1000
1 ≤ X, Y ≤ 100
1 ≤ A ≤ 100For each test case, output the minimum number of steps required by Aarav to reach Reena.<b>Sample Input</b>
4
10 20 3
36 36 5
50 4 100
30 4 2
<b>Sample Output</b>
4
0
1
13
<b>Explanation</b>
Test case 1: Here Aarav is at point 10 and Reena at psoint 20 so after waling one step Aarav will reach coordinate 13 similarly after 3 steps Aarav location will be 18. Now 3 steps Aarav have walked and only 2 steps way he is from Reena so one more step is needed of length 2. i.e. making a total of 4 steps.
Test case 2: Aarav and Reena are already at the same point., I have written this Solution Code: import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Main
{
public static void solve(){
int x = sc.nextInt();
int y = sc.nextInt();
int k = sc.nextInt();
int need = Math.abs(x-y);
int ans = (int) Math.ceil(1.0*need/k);
out.println(ans);
}
static PrintWriter out;
static Scanner sc;
public static void main (String[] args) throws java.lang.Exception
{
sc = new Scanner();
out = new PrintWriter(System.out);
int t = sc.nextInt();
while(t-->0){
solve();
}
out.close();
}
private static long power(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
private static long power(long a, long b, long m) {
a %= m;
long res = 1;
while (b > 0) {
if ((b & 1)==1)
res = res * a % m;
a = a * a % m;
b >>= 1;
}
return res;
}
private static long lcm(long a, long b) {
return a * b / gcd(a, b);
}
private static long gcd(long a, long b) {
return (b == 0) ? a : gcd(b, a % b);
}
static class Scanner {
BufferedReader br;
StringTokenizer st;
public Scanner() {
br = new BufferedReader(new
InputStreamReader(System.in));
/*
try {
br = new BufferedReader(new
InputStreamReader(new FileInputStream(new File("file_i_o\\input.txt"))));
} catch (FileNotFoundException e) {
e.printStackTrace();
}
*/
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] nextIntArray(int n) {
int[] array = new int[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
Integer[] nextIntegerArray(int n) {
Integer[] array = new Integer[n];
for (int i = 0; i < n; i++) {
array[i] = nextInt();
}
return array;
}
long[] nextLongArray(int n) {
long[] array = new long[n];
for (int i = 0; i < n; i++) {
array[i] = nextLong();
}
return array;
}
String[] nextStringArray() {
return nextLine().split(" ");
}
String[] nextStringArray(int n) {
String[] array = new String[n];
for (int i = 0; i < n; i++) {
array[i] = next();
}
return array;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Aarav is standing at location X and Reena is standing at location Y. In one step Aarav can increase or decrease his location by at most A. What is the minimum no. of steps required by Aarav to reach Reena?The first line of input will contain a single integer T, denoting the number of test cases. Each test case consists of three integers X, Y, and A, the initial coordinate of Aarav, the initial coordinate of Reena and the maximum number of coordinates Aarav can move in one step.
<b>Constraints</b>
1 ≤ T ≤1000
1 ≤ X, Y ≤ 100
1 ≤ A ≤ 100For each test case, output the minimum number of steps required by Aarav to reach Reena.<b>Sample Input</b>
4
10 20 3
36 36 5
50 4 100
30 4 2
<b>Sample Output</b>
4
0
1
13
<b>Explanation</b>
Test case 1: Here Aarav is at point 10 and Reena at psoint 20 so after waling one step Aarav will reach coordinate 13 similarly after 3 steps Aarav location will be 18. Now 3 steps Aarav have walked and only 2 steps way he is from Reena so one more step is needed of length 2. i.e. making a total of 4 steps.
Test case 2: Aarav and Reena are already at the same point., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int qw;
cin >> qw;
while (qw--) {
int p, q, cv;
cin >> p >> q >> cv;
int diff = abs(p - q);
int ans = (diff % cv) ? (diff / cv + 1) : diff / cv;
cout << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: n=int(input())
bSum=0
wSum=0
for i in range(n):
c=(input().split())
for j in range(len(c)):
if(i%2==0):
if(j%2==0):
bSum+=int(c[j])
else:
wSum+=int(c[j])
else:
if(j%2==0):
wSum+=int(c[j])
else:
bSum+=int(c[j])
print(bSum)
print(wSum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a chessboard of size N x N, where the top left square is black. Each square contains a value. Find the sum of the values of all black squares and all white squares.
Remember that in a chessboard black and white squares are alternate.The first line of input will be the N size of the matrix. Then next N lines will consist of elements of the matrix. Each row will contain N elements since it is a square matrix.
<b>Constraints:-</b>
1 ≤ N ≤ 800
1 ≤ Matrix[i][j] ≤ 100000
Print two lines, the first line containing the sum of black squares and the second line containing the sum of white squares.Input 1:
3
1 2 3
4 5 6
7 8 9
Output 1:
25
20
Sample Input 2:
4
1 2 3 4
6 8 9 10
11 12 13 14
15 16 17 18
Sample Output 2:
80
79
<b>Explanation 1</b>
The black square contains 1, 3, 5, 7, 9; sum = 25
The white square contains 2, 4, 6, 8; sum = 20, I have written this Solution Code: import java.util.*;
import java.io.*;
import java.lang.*;
class Main
{
public static void main (String[] args)throws IOException {
Scanner sc = new Scanner(System.in);
int N = sc.nextInt();
int mat[][] = new int[N][N];
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
mat[i][j] = sc.nextInt();
}
alternate_Matrix_Sum(mat,N);
}
static void alternate_Matrix_Sum(int mat[][], int N)
{
long sum =0, sum1 = 0;
for(int i = 0; i < N; i++)
{
for(int j = 0; j < N; j++)
{
if((i+j)%2 == 0)
sum += mat[i][j];
else sum1 += mat[i][j];
}
}
System.out.println(sum);
System.out.print(sum1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find the least subarray size <b>K</b> required from two arrays <b>A</b> and <b>B</b> (their positions don't matter), so that the product of the sum of those two subarrays is at least <b>S</b>. If no answer exists, print -1.The first line contains the integer T(the number of test cases).
Each test case contains 4 lines-
The first line contains 2 integers N and M (the respective sizes of the array)
The next line contains N integers (elements of the first array)
The next line contains M integers (elements of the second array)
The next line contain the integer S
Constraints:
1 <= T <= 10
1 <= N, M <= 10<sup>5</sup>
1 <= A<sub>i</sub> <= 10^4
1 <= B<sub>i</sub> <= 10^4
1 <= S <= 10<sup>16</sup>
Print the required minimum size K if a solution exists else print -1.
Print answers for each test case in a new line.
Sample Input:
1
5 6
1 2 3 4 6
2 4 5 4 9 3
120
Sample Output:
2
Explanation:
We can choose a subarray [4, 6] of size 2 from A.
We can choose a subarray [4, 9] of size 2 from B.
Now sum from array A = 4+6 = 10
Sun from array B = 4+9 = 13
Product = 10 * 13 = 130 (which is greater than S)
It can be shown that jo subarray of size 1 can give appropriate results., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL);
#define int long long
#define pb push_back
#define ff first
#define ss second
#define endl '\n'
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
using T = pair<int, int>;
typedef long double ld;
const int mod = 1e9 + 7;
const int INF = 1e9;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// template <typename T>
// using ordered_set = tree
// <T, null_type, less<T>,
// rb_tree_tag,tree_order_statistics_node_update>;
// ordered_set<pair<int, int>> s;
void solve(){
int n, m;
cin >> n >> m;
vector<int> a(n);
vector<int> b(m);
for(auto &i : a) cin >> i;
for(auto &i : b) cin >> i;
int p;
cin >> p;
int l = 1, r = min(n, m), ans = -1;
while(l <= r){
int mid = (l + r)/2;
int sum1 = 0, sum2 = 0, cur = 0;
for(int i = 0; i < n; i++){
cur += a[i];
if(i >= mid) cur -= a[i - mid];
sum1 = max(sum1, cur);
}
cur = 0;
for(int i = 0; i < m; i++){
cur += b[i];
if(i >= mid) cur -= b[i - mid];
sum2 = max(sum2, cur);
}
int temp = sum1 * sum2;
if(temp >= p){
ans = mid;
r = mid - 1;
}
else l = mid + 1;
}
cout << ans;
}
signed main(){
fast
int t = 1;
cin >> t;
for(int i = 1; i <= t; i++){
solve();
if(i != t) cout << endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array with all distinct elements which include all elements of a binary search tree in an increasing order manner (in-order traversal). But 2 element swapped their position. we have to print the sum of two swapped nodes.The first line contains a single integer N.
The second line contains N space-separated integer A[i].
Print sum of exchanged node value.Sample Input 1:
1 2 3 4 5 7 6
Sample Output 1:
13
Explanations:
if we draw a BST from given array than we can observe that element 6 and 7 had swapped their position.
So sum of these 2 is equal to 13., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main(String[] args) throws Exception {
InputStreamReader is = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(is);
int n = Integer.parseInt(br.readLine());
int[] arr = new int[n];
String[] strNums;
strNums = br.readLine().split(" ");
arr[0] = Integer.parseInt(strNums[0]);
int count = 0;
for(int i = 1; i < n; i++){
int val = Integer.parseInt(strNums[i]);
arr[i] = val;
if(arr[i-1] > arr[i]) count++;
}
int ans = 0;
if(count > 1){
int a = -1;
int b = -1;
for(int i = 1; i < n; i++){
if(arr[i-1] > arr[i] && a == -1){
a = arr[i-1];
}
else if(arr[i-1] > arr[i]){
b = arr[i];
}
}
ans = a + b;
}
else{
for(int i = 1; i < n; i++){
if(arr[i-1] > arr[i]){
ans = arr[i-1] + arr[i];
}
}
}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array with all distinct elements which include all elements of a binary search tree in an increasing order manner (in-order traversal). But 2 element swapped their position. we have to print the sum of two swapped nodes.The first line contains a single integer N.
The second line contains N space-separated integer A[i].
Print sum of exchanged node value.Sample Input 1:
1 2 3 4 5 7 6
Sample Output 1:
13
Explanations:
if we draw a BST from given array than we can observe that element 6 and 7 had swapped their position.
So sum of these 2 is equal to 13., I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int arr[n],brr[n];
for(int i=0;i<n;i++)cin>>arr[i],brr[i]=arr[i];
sort(arr,arr+n);
int sum=0;
for(int i=0;i<n;i++){
if(arr[i]!=brr[i])sum+=arr[i];
}
cout<<sum<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static class Reader{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader(){
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException{
din = new DataInputStream(new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException{
byte[] buf = new byte[64];
int cnt = 0, c;
while ((c = read()) != -1){
if (c == '\n')break;
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do{
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)return -ret;return ret;
}
public long nextLong() throws IOException{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)c = read();
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.'){
while ((c = read()) >= '0' && c <= '9'){
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)buffer[0] = -1;
}
private byte read() throws IOException{
if (bufferPointer == bytesRead)fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException{
if (din == null)return;
din.close();
}
}
public static void main (String[] args) throws IOException{
Reader sc = new Reader();
int m = sc.nextInt();
int n = sc.nextInt();
int[][] arr = new int[m][n];
for(int i = 0; i < m; i++){
for(int j = 0; j < n; j++){
arr[i][j] = sc.nextInt();
}
}
int max_row_index = 0;
int j = n - 1;
for (int i = 0; i < m; i++) {
while (j >= 0 && arr[i][j] == 1) {
j = j - 1;
max_row_index = i;
}
}
System.out.println(max_row_index);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: r, c = list(map(int, input().split()))
max_count = 0
max_r = 0
for i in range(r):
count = input().count("1")
if count > max_count:
max_count = count
max_r = i
print(max_r), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
int a[max1][max1];
signed main()
{
int n,m;
cin>>n>>m;
FOR(i,n){
FOR(j,m){cin>>a[i][j];}}
int cnt=0;
int ans=0;
int res=0;
FOR(i,n){
cnt=0;
FOR(j,m){
if(a[i][j]==1){
cnt++;
}}
if(cnt>res){
res=cnt;
ans=i;
}
}
out(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a boolean matrix of size N*M in which each row is sorted your task is to print the index of the row containing maximum 1's. If multiple answer exist print the smallest one.First line contains two space separated integers denoting values of N and M. Next N lines contains M space separated integers depicting the values of the matrix.
Constraints:-
1 < = M, N < = 1000
0 < = Matrix[][] < = 1Print the smallest index (0 indexing) of a row containing the maximum number of 1's.Sample Input:-
3 5
0 1 1 1 1
0 0 0 1 1
0 0 0 1 1
Sample Output:-
0
Sample Input:-
4 4
0 1 1 1
1 1 1 1
0 0 1 1
1 1 1 1
Sample Output:-
1, I have written this Solution Code: // mat is the matrix/ 2d array
// n,m are dimensions
function max1Row(mat, n, m) {
// write code here
// do not console.log
// return the answer as a number
let j, max_row_index = 0;
j = m - 1;
for (let i = 0; i < n; i++)
{
// Move left until a 0 is found
let flag = false;
// to check whether a row has more 1's than previous
while (j >= 0 && mat[i][j] == 1)
{
j = j - 1; // Update the index of leftmost 1
// seen so far
flag = true;//present row has more 1's than previous
}
// if the present row has more 1's than previous
if (flag)
{
max_row_index = i; // Update max_row_index
}
}
if (max_row_index == 0 && mat[0][m - 1] == 0)
return -1;
return max_row_index;
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: import java.io.*;
import java.util.*;
public class Main {
private static final int START_TEST_CASE = 1;
public static void solveCase(FastIO io, int testCase) {
final long N = io.nextLong();
int count = 0;
for (int i = 1; i <= 200; ++i) {
if (N % i == 0) {
long j = N / i;
if (digitSum(j) == i) {
++count;
}
}
}
io.println(count);
}
private static long digitSum(long x) {
long s = 0;
while (x > 0) {
s += x % 10;
x /= 10;
}
return s;
}
public static void solve(FastIO io) {
final int T = io.nextInt();
for (int t = 0; t < T; ++t) {
solveCase(io, START_TEST_CASE + t);
}
}
public static class FastIO {
private InputStream reader;
private PrintWriter writer;
private byte[] buf = new byte[1024];
private int curChar;
private int numChars;
public FastIO(InputStream r, OutputStream w) {
reader = r;
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(w)));
}
public int read() {
if (numChars == -1)
throw new InputMismatchException();
if (curChar >= numChars) {
curChar = 0;
try {
numChars = reader.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (numChars <= 0)
return -1;
}
return buf[curChar++];
}
public String nextLine() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isEndOfLine(c));
return res.toString();
}
public String nextString() {
int c = read();
while (isSpaceChar(c))
c = read();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = read();
} while (!isSpaceChar(c));
return res.toString();
}
public long nextLong() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public int nextInt() {
int c = read();
while (isSpaceChar(c))
c = read();
int sgn = 1;
if (c == '-') {
sgn = -1;
c = read();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = read();
} while (!isSpaceChar(c));
return res * sgn;
}
public double nextDouble() {
return Double.parseDouble(nextString());
}
public int[] nextIntArray(int n) {
return nextIntArray(n, 0);
}
public int[] nextIntArray(int n, int off) {
int[] arr = new int[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextInt();
}
return arr;
}
public long[] nextLongArray(int n) {
return nextLongArray(n, 0);
}
public long[] nextLongArray(int n, int off) {
long[] arr = new long[n + off];
for (int i = 0; i < n; i++) {
arr[i + off] = nextLong();
}
return arr;
}
private boolean isSpaceChar(int c) {
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(objects[i]);
}
}
public void println(Object... objects) {
print(objects);
writer.println();
}
public void printArray(int[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printArray(long[] arr) {
for (int i = 0; i < arr.length; i++) {
if (i != 0) {
writer.print(' ');
}
writer.print(arr[i]);
}
}
public void printlnArray(int[] arr) {
printArray(arr);
writer.println();
}
public void printlnArray(long[] arr) {
printArray(arr);
writer.println();
}
public void printf(String format, Object... args) {
print(String.format(format, args));
}
public void flush() {
writer.flush();
}
}
private static class Solution implements Runnable {
@Override
public void run() {
FastIO io = new FastIO(System.in, System.out);
solve(io);
io.flush();
}
}
public static void main(String[] args) throws InterruptedException {
Thread t = new Thread(null, new Solution(), "Solution", 1 << 30);
t.start();
t.join();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code:
q = int(input())
for _ in range(q):
n = int(input())
count = 0
for i in range(1,9*len(str(n))):
if not n % i:
dig = n//i
if sum(map(int,str(dig))) == i:
count += 1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alice has Q queries for you. If you can solve all her queries, she will invite you to participate in a fireworks display with her.
In each query, she gives you a positive integer N, and you have to find the number of positive integers M such that M × d(M) = N.
Here, d(M) refers to the sum of all digits of M. For example, d(1023) = 6.The first line consists of a single integer Q – the number of queries.
Then Q lines follow, each line containing a single integer N denoting a query.
<b> Constraints: </b>
1 ≤ Q ≤ 1000
1 ≤ N ≤ 10<sup>18</sup>Print Q lines, the i<sup>th</sup> line containing the answer to the i<sup>th</sup> query.Sample Input 1:
3
1
6
36
Sample Output 1:
1
0
2
Sample Explanation 1:
For the first query, the only possibility is 1.
For the third query, the only possibilities are 6 and 12., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
// #pragma gcc optimize("ofast")
// #pragma gcc target("avx,avx2,fma")
#define all(x) (x).begin(), (x).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=4e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename t>
using v = vector<t>;
template <typename t>
using vv = vector<vector<t>>;
template <typename t>
using vvv = vector<vector<vector<t>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1ll * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double pi = acosl(-1);
vector<vector<int> > multiply(vector<vector<int>> a, vector<vector<int>> b, int in_mod)
{
int n = a.size();
int l = b.size();
int m = b[0].size();
vector<vector<int> > result(n,vector<int>(n));
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
for(int k=0;k<l;k++)
{
result[i][j] = (result[i][j] + a[i][k]*b[k][j])%in_mod;
}
}
}
return result;
}
vector<vector<int>> operator%(vector<vector<int>> a, int in_mod)
{
for(auto &i:a)
for(auto &j:i)
j%=in_mod;
return a;
}
vector<vector<int>> mult_identity(vector<vector<int>> a)
{
int n=a.size();
vector<vector<int>> output(n, vector<int> (n));
for(int i=0;i<n;i++)
output[i][i]=1;
return output;
}
vector<int> mat_vector_product(vector<vector<int>> a, vector<int> b, int in_mod)
{
int n =a.size();
vector<int> output(n);
for(int i=0;i<n;i++)
{
for(int j=0;j<n;j++)
{
output[i]+=a[i][j]*b[j];
output[i]%=in_mod;
}
}
return output;
}
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % in_mod;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
int digit_sum(int x){
int sm = 0;
while(x){
sm += x%10;
x/= 10;
}
return sm;
}
void solv(){
int n;
cin>>n;
int cnt = 0;
for(int sm = 1;sm<= 200;sm++){
if(n %sm == 0){
if( digit_sum(n/sm) == sm){
cnt++;
}
}
}
cout<<cnt<<endl;
}
void solve()
{
int t = 1;
cin>>t;
for(int T=1;T<=t;T++)
{
// cout<<"Case #"<<T<<": ";
solv();
}
}
signed main()
{
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
clk = clock() - clk;
#ifndef ONLINE_JUDGE
cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}
/*
000100
1000100
1 0 -1 -2 -1 -2 -3
*/
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers find the number of mountains in the array. A mountain is a tuple (i,j,k,l,m) such that i < j < k < l < m and Arr[i] < Arr[j] < Arr[k] > Arr[l] > Arr[m].
As the number of such mountains can be large print number of mountains modulo 1000000007The first line of input contains two integers N.
Second line contains N space seperated integers.
Constraints
1 <= N <= 100000
1 <= Arr[i] <= 1000000000Output a single integer, the required value modulo 1000000007.sample input
5
1 2 3 2 1
sample output
1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define pb push_back
#define mp make_pair
#define mod 1000000007
#define For(i,n) for(int i=0;i<n;i++)
#define ff first
#define ss second
#define mem(a,b) memset(a,b,sizeof(a))
#define int long long
#define ld long double
int power_mod(int num,int g)
{
if(g==0)return 1;
if(g%2==1)return (num*power_mod((num*num)%mod,g/2))%mod;
return power_mod((num*num)%mod,g/2);
}
int power(int num,int g)
{
if(g==0)return 1;
if(g%2==1)return (num*power((num*num),g/2));
return power((num*num),g/2);
}
pair<int,int> pa[100005];
int tree[400005];
void update(int start,int end,int index,int ind,int val) {
if(start==end) {
tree[index]+=val;
return;
}
int mid=(start+end)/2;
if(ind<=mid)update(start,mid,2*index,ind,val);
else update(mid+1,end,2*index+1,ind,val);
tree[index]=tree[2*index]+tree[2*index+1];
tree[index]%=mod;
}
int search(int start,int end,int index,int l,int r) {
if(start>r || end<l)return 0;
if(start>=l && end<=r)return tree[index];
int mid=(start+end)/2;
return search(start,mid,2*index,l,r)+search(mid+1,end,2*index+1,l,r);
}
int32_t main()
{
// #ifndef ONLINE_JUDGE
// freopen("input.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
// #endif
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cout.tie(NULL);
int n;
cin>>n;
int a[n];
For(i,n)cin>>a[i];
set<int> st;
For(i,n)st.insert(a[i]);
map<int,int> mp1;
int z=1;
for(auto it=st.begin();it!=st.end();it++) {
mp1[*it]=z;
z++;
}
For(i,n)a[i]=mp1[a[i]];
mem(tree,0);
int b[n];
For(i,n) {
b[i]=search(1,z,1,1,a[i]-1);
update(1,z,1,a[i],1);
b[i]%=mod;
}
mem(tree,0);
int c[n];
For(i,n) {
c[i]=search(1,z,1,1,a[i]-1);
update(1,z,1,a[i],b[i]);
c[i]%=mod;
}
For(i,n/2)swap(a[i],a[n-i-1]);
mem(tree,0);
For(i,n) {
b[i]=search(1,z,1,1,a[i]-1);
update(1,z,1,a[i],1);
b[i]%=mod;
}
mem(tree,0);
int d[n];
For(i,n) {
d[i]=search(1,z,1,1,a[i]-1);
update(1,z,1,a[i],b[i]);
d[i]%=mod;
}
For(i,n/2)swap(d[i],d[n-i-1]);
int ans=0;
For(i,n) {
d[i]%=mod;
c[i]%=mod;
ans+=c[i]*d[i];
ans%=mod;
}
cout<<ans<<endl;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers find the number of mountains in the array. A mountain is a tuple (i,j,k,l,m) such that i < j < k < l < m and Arr[i] < Arr[j] < Arr[k] > Arr[l] > Arr[m].
As the number of such mountains can be large print number of mountains modulo 1000000007The first line of input contains two integers N.
Second line contains N space seperated integers.
Constraints
1 <= N <= 100000
1 <= Arr[i] <= 1000000000Output a single integer, the required value modulo 1000000007.sample input
5
1 2 3 2 1
sample output
1, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.awt.*;
import java.math.*;
class Main {
public static void main(String[] args) throws Exception {
new Solver().solve();
}
static class Solver {
final Helper hp;
final int MAXN = 1000_006;
final long MOD = (long) 1e9 + 7;
Solver() {
hp = new Helper(MOD, MAXN);
hp.initIO(System.in, System.out);
}
void solve() throws Exception {
int i, j, k;
FenwickTree[] seg;
TreeMap<Integer, Long>[] pending = new TreeMap[]{new TreeMap<>(), new TreeMap<>()};
long last;
int N = hp.nextInt();
if (N < 5) {
System.out.println(0);
return;
}
long[] A = hp.getLongArray(N);
Integer[] idx = new Integer[N];
for (i = 0; i < N; ++i) idx[i] = i;
Arrays.sort(idx, (a, b) -> Long.compare(A[a], A[b]));
long[][] prefix = new long[N][3];
seg = new FenwickTree[]{new FenwickTree(N), new FenwickTree(N)};
last = -7;
for (int ind : idx) {
if (A[ind] != last) {
while (!pending[0].isEmpty()) {
Map.Entry<Integer, Long> entry = pending[0].pollFirstEntry();
seg[0].update(entry.getKey(), entry.getValue());
}
while (!pending[1].isEmpty()) {
Map.Entry<Integer, Long> entry = pending[1].pollFirstEntry();
seg[1].update(entry.getKey(), entry.getValue());
}
last = A[ind];
}
++prefix[ind][0];
prefix[ind][1] = seg[0].query(0, ind - 1);
prefix[ind][2] = seg[1].query(0, ind - 1);
pending[0].put(ind, 1l);
pending[1].put(ind, prefix[ind][1]);
}
pending[0].clear();
pending[1].clear();
long[][] suffix = new long[N][3];
seg = new FenwickTree[]{new FenwickTree(N), new FenwickTree(N)};
last = -7;
for (int ind : idx) {
if (A[ind] != last) {
while (!pending[0].isEmpty()) {
Map.Entry<Integer, Long> entry = pending[0].pollFirstEntry();
seg[0].update(entry.getKey(), entry.getValue());
}
while (!pending[1].isEmpty()) {
Map.Entry<Integer, Long> entry = pending[1].pollFirstEntry();
seg[1].update(entry.getKey(), entry.getValue());
}
last = A[ind];
}
++suffix[ind][0];
suffix[ind][1] = seg[0].query(ind + 1, N - 1);
suffix[ind][2] = seg[1].query(ind + 1, N - 1);
pending[0].put(ind, 1l);
pending[1].put(ind, suffix[ind][1]);
}
pending[0].clear();
pending[1].clear();
long ans = 0;
for (i = 0; i < N; ++i) {
ans = (ans + prefix[i][2] * suffix[i][2] % MOD) % MOD;
}
hp.println(ans);
hp.flush();
}
}
static class FenwickTree {
final long MOD = (long) 1e9 + 7;
int N;
long[] tree;
FenwickTree(int size) {
N = size + 1;
tree = new long[N];
}
void update(int idx, long val) {
++idx;
while (idx < N) {
tree[idx] = (tree[idx] + val) % MOD;
idx += idx & -idx;
}
}
long query(int idx) {
++idx;
long ret = 0;
while (idx > 0) {
ret = (ret + tree[idx]) % MOD;
idx -= idx & -idx;
}
return ret;
}
long query(int l, int r) {
long ret = query(r);
if (l > 0) ret -= query(l - 1);
return (ret % MOD + MOD) % MOD;
}
}
static class Helper {
final long MOD;
final int MAXN;
final Random rnd;
public Helper(long mod, int maxn) {
MOD = mod;
MAXN = maxn;
rnd = new Random();
}
public static int[] sieve;
public static ArrayList<Integer> primes;
public void setSieve() {
primes = new ArrayList<>();
sieve = new int[MAXN];
int i, j;
for (i = 2; i < MAXN; ++i)
if (sieve[i] == 0) {
primes.add(i);
for (j = i; j < MAXN; j += i) {
sieve[j] = i;
}
}
}
public static long[] factorial;
public void setFactorial() {
factorial = new long[MAXN];
factorial[0] = 1;
for (int i = 1; i < MAXN; ++i) factorial[i] = factorial[i - 1] * i % MOD;
}
public long getFactorial(int n) {
if (factorial == null) setFactorial();
return factorial[n];
}
public long ncr(int n, int r) {
if (r > n) return 0;
if (factorial == null) setFactorial();
long numerator = factorial[n];
long denominator = factorial[r] * factorial[n - r] % MOD;
return numerator * pow(denominator, MOD - 2, MOD) % MOD;
}
public long[] getLongArray(int size) throws Exception {
long[] ar = new long[size];
for (int i = 0; i < size; ++i) ar[i] = nextLong();
return ar;
}
public int[] getIntArray(int size) throws Exception {
int[] ar = new int[size];
for (int i = 0; i < size; ++i) ar[i] = nextInt();
return ar;
}
public String[] getStringArray(int size) throws Exception {
String[] ar = new String[size];
for (int i = 0; i < size; ++i) ar[i] = next();
return ar;
}
public String joinElements(long[] ar) {
StringBuilder sb = new StringBuilder();
for (long itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(int[] ar) {
StringBuilder sb = new StringBuilder();
for (int itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(String[] ar) {
StringBuilder sb = new StringBuilder();
for (String itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public String joinElements(Object[] ar) {
StringBuilder sb = new StringBuilder();
for (Object itr : ar) sb.append(itr).append(" ");
return sb.toString().trim();
}
public long gcd(long a, long b) {
return b == 0 ? a : gcd(b, a % b);
}
public int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a % b);
}
public long max(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.max(ret, itr);
return ret;
}
public int max(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.max(ret, itr);
return ret;
}
public long min(long[] ar) {
long ret = ar[0];
for (long itr : ar) ret = Math.min(ret, itr);
return ret;
}
public int min(int[] ar) {
int ret = ar[0];
for (int itr : ar) ret = Math.min(ret, itr);
return ret;
}
public long sum(long[] ar) {
long sum = 0;
for (long itr : ar) sum += itr;
return sum;
}
public long sum(int[] ar) {
long sum = 0;
for (int itr : ar) sum += itr;
return sum;
}
public void shuffle(int[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public void shuffle(long[] ar) {
int r;
for (int i = 0; i < ar.length; ++i) {
r = rnd.nextInt(ar.length);
if (r != i) {
ar[i] ^= ar[r];
ar[r] ^= ar[i];
ar[i] ^= ar[r];
}
}
}
public long pow(long base, long exp, long MOD) {
base %= MOD;
long ret = 1;
while (exp > 0) {
if ((exp & 1) == 1) ret = ret * base % MOD;
base = base * base % MOD;
exp >>= 1;
}
return ret;
}
static byte[] buf = new byte[2048];
static int index, total;
static InputStream in;
static BufferedWriter bw;
public void initIO(InputStream is, OutputStream os) {
try {
in = is;
bw = new BufferedWriter(new OutputStreamWriter(os));
} catch (Exception e) {
}
}
public void initIO(String inputFile, String outputFile) {
try {
in = new FileInputStream(inputFile);
bw = new BufferedWriter(new OutputStreamWriter(
new FileOutputStream(outputFile)));
} catch (Exception e) {
}
}
private int scan() throws Exception {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0)
return -1;
}
return buf[index++];
}
public String next() throws Exception {
int c;
for (c = scan(); c <= 32; c = scan()) ;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan())
sb.append((char) c);
return sb.toString();
}
public int nextInt() throws Exception {
int c, val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public long nextLong() throws Exception {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan()) ;
boolean neg = c == '-';
if (c == '-' || c == '+')
c = scan();
for (; c >= '0' && c <= '9'; c = scan())
val = (val << 3) + (val << 1) + (c & 15);
return neg ? -val : val;
}
public void print(Object a) throws Exception {
bw.write(a.toString());
}
public void printsp(Object a) throws Exception {
print(a);
print(" ");
}
public void println() throws Exception {
bw.write("\n");
}
public void println(Object a) throws Exception {
print(a);
println();
}
public void flush() throws Exception {
bw.flush();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: static void verticalFive(){
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
System.out.println("*");
}
static void horizontalFive(){
System.out.print("* * * * *");
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print Five stars ('*') <b><i>vertically</i></b> and 5 <b><i>horizontally</i></b>
There will be two functions:
<ul>
<li>verticalFive(): Print stars in vertical order</li>
<li>horizontalFive(): Print stars in horizontal order</l>
</ul><b>User Task:</b>
Your task is to complete the functions <b>verticalFive()</b> and <b>horizontalFive()</b>.
Print 5 vertical stars in <b> verticalFive</b> and 5 horizontal stars(separated by whitespace) in <b>horizontalFive</b> function.
<b>Note</b>: You don't need to print the extra blank line it will be printed by the driver codeNo Sample Input:
Sample Output:
*
*
*
*
*
* * * * *, I have written this Solution Code: def vertical5():
for i in range(0,5):
print("*",end="\n")
#print()
def horizontal5():
for i in range(0,5):
print("*",end=" ")
vertical5()
print(end="\n")
horizontal5(), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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