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For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: def checkConevrtion(a):
return str(a)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer <b>N</b>, you need to typecast this integer to String. If the typecasting is done successfully then we will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".User task:
Since this is a functional problem you don't have to worry about the input. You just have to complete the function <b>checkConvertion()</b>, which contains N as a parameter.You need to return the typecasted string value. The driver code will print "<b>Nice Job</b>" otherwise "<b>Wrong answer</b>".Sample Input:
5
Sample Output:
Nice Job
Sample Input:
6
Sample Output:
Nice Job, I have written this Solution Code: static String checkConevrtion(int a)
{
return String.valueOf(a);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number 'N'. The task is to find the Nth number whose each digit is a prime number(<10) i.e 2, 3, 5, 7. In other words you have to find nth number of this sequence : 2, 3, 5, 7, 22, 23 ,.. and so on.The first line contains a single integer T i.e. the number of test cases. The first and only line of each test case consists of a single integer N.
Constraints:
1 <= T <= 100
1 <= N <= 100000For each testcase print the nth number of the given sequence made of prime digits in a new line.Input:
2
10
21
Output:
33
222
Explanation:
Testcase 1: 10th number in the sequence of numbers whose each digit is prime is 33.
Testcase 2: 21th number in the sequence of numbers whose each digit is prime is 222., 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());
while(t-->0)
{
int n = Integer.parseInt(br.readLine());
String ans = check(n);
StringBuffer sbr = new StringBuffer(ans);
System.out.println(sbr.reverse());
}
}
static String check(int x)
{
String ans="";
while(x>0)
{
switch(x%4)
{
case 1:ans +="2"; break;
case 2:ans +="3"; break;
case 3:ans+="5"; break;
case 0:ans+="7"; break;
}
if(x%4==0)
x--;
x/=4;
}
return ans;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number 'N'. The task is to find the Nth number whose each digit is a prime number(<10) i.e 2, 3, 5, 7. In other words you have to find nth number of this sequence : 2, 3, 5, 7, 22, 23 ,.. and so on.The first line contains a single integer T i.e. the number of test cases. The first and only line of each test case consists of a single integer N.
Constraints:
1 <= T <= 100
1 <= N <= 100000For each testcase print the nth number of the given sequence made of prime digits in a new line.Input:
2
10
21
Output:
33
222
Explanation:
Testcase 1: 10th number in the sequence of numbers whose each digit is prime is 33.
Testcase 2: 21th number in the sequence of numbers whose each digit is prime is 222., I have written this Solution Code: def nthprimedigitsnumber(number):
num=""
while (number > 0):
rem = number % 4
if (rem == 1):
num += '2'
if (rem == 2):
num += '3'
if (rem == 3):
num += '5'
if (rem == 0):
num += '7'
if (number % 4 == 0):
number = number - 1
number = number // 4
return num[::-1]
T=int(input())
for i in range(T):
number = int(input())
print(nthprimedigitsnumber(number)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number 'N'. The task is to find the Nth number whose each digit is a prime number(<10) i.e 2, 3, 5, 7. In other words you have to find nth number of this sequence : 2, 3, 5, 7, 22, 23 ,.. and so on.The first line contains a single integer T i.e. the number of test cases. The first and only line of each test case consists of a single integer N.
Constraints:
1 <= T <= 100
1 <= N <= 100000For each testcase print the nth number of the given sequence made of prime digits in a new line.Input:
2
10
21
Output:
33
222
Explanation:
Testcase 1: 10th number in the sequence of numbers whose each digit is prime is 33.
Testcase 2: 21th number in the sequence of numbers whose each digit is prime is 222., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
void nthprimedigitsnumber(long long n)
{
long long len = 1;
long long prev_count = 0;
while (true) {
long long curr_count = prev_count + pow(4, len);
if (prev_count < n && curr_count >= n)
break;
len++;
prev_count = curr_count;
}
for (int i = 1; i <= len; i++) {
for (long long j = 1; j <= 4; j++) {
if (prev_count + pow(4, len - i) < n)
prev_count += pow(4, len - i);
else {
if (j == 1)
cout << "2";
else if (j == 2)
cout << "3";
else if (j == 3)
cout << "5";
else if (j == 4)
cout << "7";
break;
}
}
}
cout << endl;
}
int main(){
int t;
cin>>t;
while(t--){
long long n;
cin>>n;
nthprimedigitsnumber(n);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has two types of apples that have some sweetness level. She has N apples of the first type which costs her P rupees each and M apples of the second type which costs her Q rupees each. She wants to pick apples in such a way that the total sum of sweetness level is equal to or more than X and the total cost is minimum.The first line of input contains 5 integers N, M, P, Q, X separated by spaces. The second line of input contains N Space- separated integers depicting the sweetness level of the first type of apples. The last line of input contains M space- separated integers depicting the sweetness level of the second type of apples.
Sample Input:-
1 <= N, M <= 50000
1 <= X, P, Q <= 1000000000
1 <= Sweetness <= 100000Print the minimum cost required if it is possible to make total sweetness more than X else print -1.Sample Input:-
5 5 1 2 16
1 2 3 4 5
3 2 1 8 7
Sample Output:-
4
Explanation:-
From first type:- 4, 5
From second type:- 8
Sample Input:-
5 5 1 2 100
1 2 3 4 5
1 2 3 4 5
Sample Output:-
-1, I have written this Solution Code: import sys
n,m,p,q,x=map(int,input().split())
a=list(map(int,input().split()))
b=list(map(int,input().split()))
a.sort(reverse=True)
b.sort()
t=sum(b)
v=sys.maxsize
c=q*m
i=j=0
while i<m and j<n:
while t>=x and i<m:
t-=b[i]
i+=1
c-=q
if t>=x and c<v:v=c
while t<x and j<n:
t+=a[j]
j+=1
c+=p
if t>=x and c<v:v=c
if v==sys.maxsize:print(-1)
else:print(v), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has two types of apples that have some sweetness level. She has N apples of the first type which costs her P rupees each and M apples of the second type which costs her Q rupees each. She wants to pick apples in such a way that the total sum of sweetness level is equal to or more than X and the total cost is minimum.The first line of input contains 5 integers N, M, P, Q, X separated by spaces. The second line of input contains N Space- separated integers depicting the sweetness level of the first type of apples. The last line of input contains M space- separated integers depicting the sweetness level of the second type of apples.
Sample Input:-
1 <= N, M <= 50000
1 <= X, P, Q <= 1000000000
1 <= Sweetness <= 100000Print the minimum cost required if it is possible to make total sweetness more than X else print -1.Sample Input:-
5 5 1 2 16
1 2 3 4 5
3 2 1 8 7
Sample Output:-
4
Explanation:-
From first type:- 4, 5
From second type:- 8
Sample Input:-
5 5 1 2 100
1 2 3 4 5
1 2 3 4 5
Sample Output:-
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int t;
t=1;
while(t--){
int n,m;
int l,g,x;
cin>>n>>m>>l>>g>>x;
vector<int> a,b;
int q;
for(int i=0;i<n;i++){
cin>> q;
b.emplace_back(q);
}
for(int i=0;i<m;i++){
cin>>q;
a.emplace_back(q);
}
sort(a.begin(),a.end());
sort(b.begin(),b.end());
reverse(a.begin(),a.end());
reverse(b.begin(),b.end());
int sum=0;
int ans=0;
n=a.size();
int p=b.size();
vector<int> prea(n+1),preb(p+1);
prea[0]=0;
preb[0]=0;
for(int i=1;i<=n;i++){
sum+=a[i-1];
prea[i]=sum;
}
sum=0;
for(int i=1;i<=p;i++){
sum+=b[i-1];
preb[i]=sum;
}
ans=LLONG_MAX;
vector<int>::iterator it;
// int x;
for(int i=0;i<=n;i++){
// cout<<prea[i]<<" ";
if(x>=prea[i]){
it=lower_bound(preb.begin(),preb.end(),x-prea[i]);
if(it!=preb.end()){
// cout<<*it<<" 2343";
ans=min(ans,(it-preb.begin())*l+g*i);
}
}
else{
ans=min(g*i,ans);
}
}
if(ans==LLONG_MAX){cout<<-1<<endl;}
else{
cout<<ans<<endl;
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara has two types of apples that have some sweetness level. She has N apples of the first type which costs her P rupees each and M apples of the second type which costs her Q rupees each. She wants to pick apples in such a way that the total sum of sweetness level is equal to or more than X and the total cost is minimum.The first line of input contains 5 integers N, M, P, Q, X separated by spaces. The second line of input contains N Space- separated integers depicting the sweetness level of the first type of apples. The last line of input contains M space- separated integers depicting the sweetness level of the second type of apples.
Sample Input:-
1 <= N, M <= 50000
1 <= X, P, Q <= 1000000000
1 <= Sweetness <= 100000Print the minimum cost required if it is possible to make total sweetness more than X else print -1.Sample Input:-
5 5 1 2 16
1 2 3 4 5
3 2 1 8 7
Sample Output:-
4
Explanation:-
From first type:- 4, 5
From second type:- 8
Sample Input:-
5 5 1 2 100
1 2 3 4 5
1 2 3 4 5
Sample Output:-
-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 m = sc.nextInt();
int p = sc.nextInt();
int q = sc.nextInt();
int x = sc.nextInt();
int a[] = new int[n];
int b[] = new int[m];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
for(int i =0;i<m;i++){b[i]=sc.nextInt();
}
Arrays.sort(a);
Arrays.sort(b);
long pre[]= new long[n];
long suf[] = new long[m];
pre[0]=a[n-1];
for(int i=1;i<n;i++){
pre[i]=pre[i-1]+a[n-i-1];
}
suf[0]=b[m-1];
for(int i=1;i<m;i++){
suf[i]=suf[i-1]+b[m-i-1];
}
long ans=100000000000000L;
if(suf[m-1]>=x){
ans=Math.min(ans,q*m);
}
if(pre[n-1]>=x){
ans=Math.min(ans,p*n);
}
int j=m-1;
for(int i=0;i<m;i++){
if(suf[i]>=x){j=i;
ans=Math.min(ans,q*(j+1));
break;}
}
for(int i=0;i<n;i++){
if(pre[i]>=x){ans=Math.min(ans,p*(i+1));break;}
while(j >=0 && pre[i] + suf[j]>=x){
j--;
}
if(j!=m-1){j++;}
if(pre[i]+suf[j]>=x){
ans=Math.min(ans,p*(i+1)+q*(j+1));}
}
if(ans==100000000000000L){System.out.print(-1);return ;}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
public static int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;break;}
ch*=(long)3;
}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
int Ifpossible(long x, long y){
long sum=y-x;
long ans=0;
long ch = 1;
while(ans<sum){
ans+=ch;
if(ans==sum){return 1;}
ch*=3;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two integers x and y, check if x can be converted to y by adding F(n) to x where F(n){n>=1} is defined as:- 1+3+9+27+81+. .3^n.<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>Ifpossible()</b> that takes the integer x and y as parameter.
Constraints:-
1<= x < y <= 10^9Return 1 if it is possible to convert x into y else return 0.Sample Input:-
5 7
Sample Output:-
0
Sample Input:-
3 16
Sample Output:-
1
Explanation:-
F(3) = 1 + 3 + 9 = 13
3 + 13 = 16, I have written this Solution Code:
def Ifpossible(x,y) :
result = y-x
ans = 0
ch = 1
while ans<result :
ans+=ch
if ans==result:
return 1;
ch*=3
return 0;
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: def Race(A,B,C):
if abs(C-A) ==abs(C-B):
return 'D'
if abs(C-A)>abs(C-B):
return 'S'
return 'N'
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code:
char Race(int A, int B, int C){
if(abs(C-A)==abs(C-B)){return 'D';}
if(abs(C-A)>abs(C-B)){return 'S';}
else{
return 'N';}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Naruto and Sasuke are on a S- Rank mission. Now They got bored and thought of competing in a race against each other in a horizontal plane. They decides a common spot C at which both of them will try to reach. Whoever reaches first wins the race, both of them run at the same speed.
Given initial positions of Naruto and Sasuke as A and B recpectively.
you need to tell which of them will win the race. If Naruto wins print "N" ( without the quotes ), if Sasuke wins print "S" ( without the quotes ).
if both of them reach the common spot at the same time, print "D" (for draw, without the quotes ).<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>Race</b> that takes the integer A (initial position of Naruto) and B (initial position of Sasuke) and C (position of common spot) as parameter.
Constraints
1 <= A, B, C <= 100Return the character according to the given scenario.Sample Input
1 2 3
Sample Output
S
Sample Input
1 3 2
Sample Output
D, I have written this Solution Code: static char Race(int A,int B,int C){
if(Math.abs(C-A)==Math.abs(C-B)){return 'D';}
if(Math.abs(C-A)>Math.abs(C-B)){return 'S';}
else{
return 'N';}
}, 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: 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 a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<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>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: // n is the input number
function recSum(n) {
// write code here
// do not console.log
// return the answer as a number
if (n < 10) return n;
return n % 10 + recSum(Math.floor(n / 10))
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N, find the sum of all the digits of the number
Note: Use a recursive method to solve this problem.<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>Sum()</b> that takes the integer N as a parameter.
Constraints:
1 ≤ T ≤ 100
0 ≤ N ≤ 1000000000Return sum of digits.Sample Input
2
25
28
Sample Output
7
10, I have written this Solution Code: static long Sum(long n)
{ if(n==0){return 0;}
return n%10+Sum(n/10);
} , In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<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>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: static int LikesBoth(int N, int A, int B){
return (A+B-N);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<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>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: def LikesBoth(N,A,B):
return (A+B-N)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<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>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: N people are standing in a queue in which A of them like apple and B of them like oranges. How many people like both apple and oranges.
<b>Note</b>:- It is guaranteed that each and every person likes at least one of the given two.<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>LikesBoth()</b> that takes integers N, A, and B as arguments.
Constraints:-
1 <= N <= 10000
1 <= A <= N
1 <= B <= NReturn the number of people that like both of the fruit.Sample Input:-
5 3 4
Sample Output:-
2
Sample Input:-
5 5 5
Sample Output:-
5, I have written this Solution Code: int LikesBoth(int N,int A, int B){
return (A+B-N);
}, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: function easySorting(arr)
{
for(let i = 1; i < 5; i++)
{
let str = arr[i];
let j = i-1;
while(j >= 0 && (arr[j].toString().localeCompare(str)) > 0 )
{
arr[j+1] = arr[j];
j--;
}
arr[j+1] = str;
}
return arr;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
map<string,int> m;
string s;
for(int i=0;i<5;i++){
cin>>s;
m[s]++;
}
for(auto it=m.begin();it!=m.end();it++){
while(it->second>0){
cout<<it->first<<" ";
it->second--;}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: inp = input("").split(" ")
print(" ".join(sorted(inp))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: An array of 5 string is given where each string contains 2 characters, Now you have to sort these strings, like in a dictionary.Input contains 5 strings of length 2 separated by spaces.
String contains only uppercase English letters.Print the sorted array.INPUT :
AS KF ER DD JK
OUTPUT :
AS DD ER JK KF, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
static void printArray(String str[])
{
for (String string : str)
System.out.print(string + " ");
}
public static void main (String[] args) throws IOException {
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
int len = 5;
String[] str = new String[len];
str = br.readLine().split(" ");
Arrays.sort(str, String.CASE_INSENSITIVE_ORDER);
printArray(str);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, 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));
BufferedWriter bo=new BufferedWriter(new OutputStreamWriter(System.out));
int t;
try{
t=Integer.parseInt(br.readLine());
}
catch(Exception e)
{
return;
}
while(t-->0)
{
String[] g=br.readLine().split(" ");
int n=Integer.parseInt(g[0]);
int k=Integer.parseInt(g[1]);
if(k>n || (k==1) || (k>26))
{
if(n==1 && k==1)
bo.write("a\n");
else
bo.write(-1+"\n");
}
else
{
int extra=k-2;
boolean check=true;
while(n>extra)
{
if(check==true)
bo.write("a");
else
bo.write("b");
if(check==true)
check=false;
else
check=true;
n--;
}
for(int i=0;i<extra;i++)
bo.write((char)(i+99));
bo.write("\n");
}
}
bo.close();
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: t=int(input())
for tt in range(t):
n,k=map(int,input().split())
if (k==1 and n>1) or (k>n):
print(-1)
continue
s="abcdefghijklmnopqrstuvwxyz"
ss="ab"
if (n-k)%2==0:
a=ss*((n-k)//2)+s[:k]
else:
a=ss*((n-k)//2)+s[:2]+"a"+s[2:k]
print(a), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given N and K, find the lexicographically smallest string of length N using only the first K lowercase letters of the alphabet such that each letter is used at least once and no two adjacent characters are equal.
If such a string doesn't exist, print -1.The first line of input contains a single integer, T (1 <= T <= 100).
Then T lines follow, each containing two space-separated integers, N (1 <= N <= 10<sup>5</sup>) and K (1 <= K <= 26).
It is guaranteed that sum of N over all test cases does not exceed 10<sup>6</sup>For each test case, output its answer in a new line.Sample Input:
2
2 3
3 2
Sample Output:
-1
aba, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
typedef long long int ll;
typedef unsigned long long int ull;
const long double PI = acos(-1);
const ll mod=1e9+7;
const ll mod1=998244353;
const int inf = 1e9;
const ll INF=1e18;
void precompute(){
}
void TEST_CASE(){
int n,k;
cin >> n >> k;
if(k==1){
if(n>1){
cout << -1 << endl;
}else{
cout << 'a' << endl;
}
}else if(n<k){
cout << -1 << endl;
}else if(n==k){
string s="";
for(int i=0 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}else{
string s="";
for(int i=0 ; i<(n-k+2) ; i++){
if(i%2){
s+="b";
}else{
s+="a";
}
}
for(int i=2 ; i<k ; i++){
s+=('a'+i);
}
cout << s << endl;
}
}
signed main(){
fast;
//freopen ("INPUT.txt","r",stdin);
//freopen ("OUTPUT.txt","w",stdout);
int test=1,TEST=1;
precompute();
cin >> test;
while(test--){
TEST_CASE();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., 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 t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int[] a=new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int count=search(a,0,n-1);
System.out.println(count);
}
}
public static int search(int[] a,int l,int h){
while(l<=h){
int mid=l+(h-l)/2;
if ((mid==h||a[mid+1]==0)&&(a[mid]==1))
return mid+1;
if (a[mid]==1)
l=mid+1;
else h=mid-1;
}
return 0;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., 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 = 1e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
int l = 0, h = n+1;
while(l+1 < h){
int m = (l + h) >> 1;
if(a[m] == 1)
l = m;
else
h = m;
}
cout << l << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary sorted non-increasing array arr of size <b>N</b>. You need to print the count of <b>1's</b> in the binary array.
Try to solve the problem using binary searchThe input line contains T, denotes the number of testcases.
Each test case contains two lines. The first line contains N (size of binary array). The second line contains N elements of binary array separated by space.
<b>Constraints:</b>
1 <= T <= 100
1 <= N <= 10^6
arr[i] = 0,1
<b>Sum of N over all testcases does not exceed 10^6</b>For each testcase in new line, print the count 1's in binary array.Input:
2
8
1 1 1 1 1 0 0 0
8
1 1 0 0 0 0 0 0
Output:
5
2
Explanation:
Testcase 1: Number of 1's in given binary array : 1 1 1 1 1 0 0 0 is 5.
Testcase 2: Number of 1's in given binary array : 1 1 0 0 0 0 0 0 is 2., I have written this Solution Code: c=int(input())
for x in range(c):
size=int(input())
s=input()
print(s.count('1')), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to find the Nth Catalan number.
The first few Catalan numbers for N = 1, 2, 3, … are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
You can read more about Catalan numbers <a href="https://en.wikipedia.org/wiki/Catalan_number">here</a>.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N.
Constraints:
1 <= T <= 100000
1 <= N <= 1000000For each test case, in a new line print the Catalan number at position N.
Since the answer can be large, print answer modulo (10^9 + 7)Sample Input:
3
5
4
10
Sample Output:
42
14
16796, I have written this Solution Code: import java.io.*;
class Main {
public static void main(String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
dp[0]=0;
dp[1]=1;
for (int i = 1; i <=1000000 ; i++) {
dp[i+1]=(((4*i+2)%mod)*dp[i]%mod*modPow(i+2,mod-2))%mod;
}
while(t-->0){
int n = Integer.parseInt(br.readLine());
System.out.println(dp[n]);
}
}
static long [] dp= new long [1000002];
static long mod = 1000000007;
static long modPow(long x, long n){
if(n==0) return 1;
if(n==1) return x;
if(n%2==0) return modPow((x*x)%mod,n/2)%mod;
return (x*modPow((x*x)%mod,n/2)%mod);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number N. The task is to find the Nth Catalan number.
The first few Catalan numbers for N = 1, 2, 3, … are 1, 2, 5, 14, 42, 132, 429, 1430, 4862, …
You can read more about Catalan numbers <a href="https://en.wikipedia.org/wiki/Catalan_number">here</a>.The first line of input contains a single integer T which denotes the number of test cases. The first line of each test case contains a single integer N.
Constraints:
1 <= T <= 100000
1 <= N <= 1000000For each test case, in a new line print the Catalan number at position N.
Since the answer can be large, print answer modulo (10^9 + 7)Sample Input:
3
5
4
10
Sample Output:
42
14
16796, I have written this Solution Code: #include<bits/stdc++.h>
#define int long long
#define ld long double
#define ll long long
#define pb push_back
#define endl '\n'
#define pi pair<int,int>
#define vi vector<int>
#define all(a) (a).begin(),(a).end()
#define fi first
#define se 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 = 2e6 + 5;
const int mod = 1e9 + 7;
const int inf = 1e9 + 9;
int f[N], inv[N], res[N];
int power(int a, int b){
int ans = 1;
while(b){
if(b&1)
ans = (ans*a) % mod;
b >>= 1;
a = (a*a) % mod;
}
return ans;
}
void solve(){
int n; cin >> n;
cout << res[n] << endl;
}
void testcases(){
int tt = 1;
f[0] = 1;
for(int i = 1; i < N; i++)
f[i] = (i*f[i-1]) % mod;
inv[N-1] = power(f[N-1], mod-2);
for(int i = N-2; i >= 1; i--)
inv[i] = ((i+1)*inv[i+1]) % mod;
for(int i = 1; i < N/2; i++){
res[i] = f[2*i];
res[i] = (res[i]*inv[i]) % mod;
res[i] = (res[i]*inv[i]) % mod;
res[i] = (res[i]*power(i+1, mod-2)) % mod;
}
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: Given an array <b>nums</b> of N integers where each element is in between 1 to N. Your task is to print all the integers from 1 to N which are not present in the given array.First line contains n the length of input
Second line of input contains the array of integers in range from (1 to n)
<b> Constraints </b>
1 <= n <= 100000
1 <= arr[i] <= nPrint all the integers in range [1, n] that dose not appear in array.Sample Input :
5
1 2 3 4 4
Sample Output :
5
Sample Input:
7
4 1 3 2 5 5 4
Sample Output:
6 7
Explaination:
In the first test, only 5 is the integer from 1 to n, which is missing from the array.
In the second test, both 6 and 7 are missing., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc=new Scanner(System.in);
int n=sc.nextInt();
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
List<Integer> ans=missing(arr);
for(int i=0;i<ans.size();i++){
System.out.print(ans.get(i)+" ");
}
System.out.println();
}
static List<Integer> missing(int[] arr){
int i=0;
while (i< arr.length){
int correct=arr[i]-1;
if (arr[i]!=arr[correct]){
swap(arr,i,correct);
}else {
i++;
}
}
List<Integer> ans=new ArrayList<>();
for (int j = 0; j < arr.length; j++) {
if (arr[j] != j+1) {
ans.add(j + 1);
}
}
return ans;
}
static void swap(int[] arr,int first,int second){
int temp=arr[first];
arr[first]=arr[second];
arr[second]=temp;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array <b>nums</b> of N integers where each element is in between 1 to N. Your task is to print all the integers from 1 to N which are not present in the given array.First line contains n the length of input
Second line of input contains the array of integers in range from (1 to n)
<b> Constraints </b>
1 <= n <= 100000
1 <= arr[i] <= nPrint all the integers in range [1, n] that dose not appear in array.Sample Input :
5
1 2 3 4 4
Sample Output :
5
Sample Input:
7
4 1 3 2 5 5 4
Sample Output:
6 7
Explaination:
In the first test, only 5 is the integer from 1 to n, which is missing from the array.
In the second test, both 6 and 7 are missing., I have written this Solution Code: /**
* Author : tourist1256
* Time : 2022-01-08 12:34:09
**/
#include <bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#define debug(...) cerr << "[" << #__VA_ARGS__ << "]:", debug_out(__VA_ARGS__)
#else
#define debug(...) 2351
#endif
vector<int> findDisappearedNumbers(vector<int>& nums) {
int i = 0, size = nums.size();
while (i < size) {
if (nums[i] != i + 1 && nums[nums[i] - 1] != nums[i])
swap(nums[i], nums[nums[i] - 1]);
else
i++;
}
vector<int> ans;
for (i = 0; i < size; i++)
if (nums[i] != i + 1) ans.push_back(i + 1);
return ans;
}
int main() {
int n;
cin >> n;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
vector<int> x = findDisappearedNumbers(a);
for (auto& it : x) {
cout << it << " ";
}
cout << "\n";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A[] of size N. For any subarray size i (1 <= i <= N), you have to do the following:
<ul><li>Write all subarrays of size i from array A on a piece of paper.</li><li>Now, replace each of these subarrays with their minimum element.</li><li>Now, find the maximum value among all these minimums.</li>
Print this maximum value for all subarray sizes k (1 <= k <= N).The first line contains an integer N denoting the size of the array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraint:-
1 <= N <= 10^5
1 <= A[i] <= 10^6Print the array of numbers of size N for each of the considered window size 1, 2, ..., N respectively.Sample Input:
7
10 20 30 50 10 70 30
Sample Output:
70 30 20 10 10 10 10
Explanation:
Testcase 1:
First element in output indicates maximum of minimums of all windows of size 1. Minimums of windows of size 1 are {10}, {20}, {30}, {50}, {10}, {70} and {30}. Maximum of these minimums is 70.
Second element in output indicates maximum of minimums of all windows of size 2. Minimums of windows of size 2 are {10}, {20}, {30}, {10}, {10}, and {30}. Maximum of these minimums is 30.
Third element in output indicates maximum of minimums of all windows of size 3. Minimums of windows of size 3 are {10}, {20}, {10}, {10} and {10}. Maximum of these minimums is 20.
Similarly other elements of output are computed., 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();
String[] str = s.split(" ");
int arr[] = new int[n];
for(int i=0 ; i<n ; i++){
arr[i]=Integer.parseInt(str[i]) ;
}
printMaxOfMin(n,arr);
}
static void printMaxOfMin(int n,int[] arr)
{
Stack<Integer> s = new Stack<>();
int left[] = new int[n+1];
int right[] = new int[n+1];
for (int i=0; i<n; i++)
{
left[i] = -1;
right[i] = n;
}
for (int i=0; i<n; i++)
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();
if (!s.empty())
left[i] = s.peek();
s.push(i);
}
while (!s.empty())
s.pop();
for (int i = n-1 ; i>=0 ; i-- )
{
while (!s.empty() && arr[s.peek()] >= arr[i])
s.pop();
if(!s.empty())
right[i] = s.peek();
s.push(i);
}
int ans[] = new int[n+1];
for (int i=0; i<=n; i++)
ans[i] = 0;
for (int i=0; i<n; i++)
{
int len = right[i] - left[i] - 1;
ans[len] = Math.max(ans[len], arr[i]);
}
for (int i=n-1; i>=1; i--)
ans[i] = Math.max(ans[i], ans[i+1]);
for (int i=1; i<=n; i++)
System.out.print(ans[i] + " ");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer array A[] of size N. For any subarray size i (1 <= i <= N), you have to do the following:
<ul><li>Write all subarrays of size i from array A on a piece of paper.</li><li>Now, replace each of these subarrays with their minimum element.</li><li>Now, find the maximum value among all these minimums.</li>
Print this maximum value for all subarray sizes k (1 <= k <= N).The first line contains an integer N denoting the size of the array. The second line contains N space-separated integers A1, A2, ..., AN denoting the elements of the array.
Constraint:-
1 <= N <= 10^5
1 <= A[i] <= 10^6Print the array of numbers of size N for each of the considered window size 1, 2, ..., N respectively.Sample Input:
7
10 20 30 50 10 70 30
Sample Output:
70 30 20 10 10 10 10
Explanation:
Testcase 1:
First element in output indicates maximum of minimums of all windows of size 1. Minimums of windows of size 1 are {10}, {20}, {30}, {50}, {10}, {70} and {30}. Maximum of these minimums is 70.
Second element in output indicates maximum of minimums of all windows of size 2. Minimums of windows of size 2 are {10}, {20}, {30}, {10}, {10}, and {30}. Maximum of these minimums is 30.
Third element in output indicates maximum of minimums of all windows of size 3. Minimums of windows of size 3 are {10}, {20}, {10}, {10} and {10}. Maximum of these minimums is 20.
Similarly other elements of output are computed., 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;
int l[N], r[N], ans[N], a[N];
signed main() {
IOS;
int n; cin >> n;
stack<int> s;
a[0] = a[n+1] = -inf;
s.push(0);
for(int i = 1; i <= n; i++){
cin >> a[i];
while(!s.empty() && a[s.top()] >= a[i])
s.pop();
l[i] = s.top();
s.push(i);
}
while(!s.empty())
s.pop();
s.push(n+1);
for(int i = n; i >= 1; i--){
while(!s.empty() && a[s.top()] >= a[i])
s.pop();
r[i] = s.top();
s.push(i);
}
for(int i = 1; i <= n; i++){
int len = r[i] - l[i] - 1;
ans[len] = max(ans[len], a[i]);
}
for(int i = n-1; i >= 1; i--)
ans[i] = max(ans[i], ans[i+1]);
for(int i = 1; i <= n; i++)
cout << ans[i] << " ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are n students in a class. The teacher gives x chocolates to be distributed among the students. The goodness of the class is the bitwise OR of the number of students in the class and the number of chocolates the teacher has given. Find the goodness of the class.The input consists of two space separated integers n and x.
<b>Constraints</b>
1 ≤ n ≤ 10<sup>5</sup>
1 ≤ x ≤ 10<sup>5</sup>Print the goodness of the class.<b>Sample Input 1</b>
100 5
<b>Sample Output 1</b>
101
<b>Sample Input 1</b>
64 7
<b>Sample Output 1</b>
71, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
using namespace std;
#define int long long int
#define endl '\n'
void fastio()
{
ios_base::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
using namespace __gnu_pbds;
const int M = 1e9 + 7;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> pbds;
// Author:Abhas
void solution()
{
int n, x;
cin >> n >> x;
int ans = n | x;
cout << ans << endl;
}
signed main()
{
fastio();
int t = 1;
// cin >> t;
while (t--)
{
solution();
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bruce Banner is working on a code involving 3 numbers with their sum as a mystery number. Just then the emergency alarm is sounded and Bruce hulks his way out to save the city of New York yet again. The responsibility of cracking the code is then handed over to you since it is crucial to ensure Hulk’s safety. You find the mystery number as X. Now give Hulk a number of all the triplets and help Hulk as he fights the evil Chitauri.First line of input contains a single integer N (length of the array)
second line contain the array elements
third line contain a single integer K(sum)
Constraint:-
1<=N<=1000
1<=elements<=100000
1<=K<=100000Output a single line containing the number of required tripletsSample Input:
6
1 2 3 4 5 6
8
Sample Output:-
2
Explanation:-
(1,2,5) , (1,3,4) are the required triplets
, I have written this Solution Code: t=int(input())
m=list(map(int,input().rstrip().split()))
n=int(input())
count=0
for i in range(t-1):
s=dict()
for j in range(i+1,t):
x=n-(m[i]+m[j])
if x in s.keys():
count+=1
else:
s[m[j]]=1
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bruce Banner is working on a code involving 3 numbers with their sum as a mystery number. Just then the emergency alarm is sounded and Bruce hulks his way out to save the city of New York yet again. The responsibility of cracking the code is then handed over to you since it is crucial to ensure Hulk’s safety. You find the mystery number as X. Now give Hulk a number of all the triplets and help Hulk as he fights the evil Chitauri.First line of input contains a single integer N (length of the array)
second line contain the array elements
third line contain a single integer K(sum)
Constraint:-
1<=N<=1000
1<=elements<=100000
1<=K<=100000Output a single line containing the number of required tripletsSample Input:
6
1 2 3 4 5 6
8
Sample Output:-
2
Explanation:-
(1,2,5) , (1,3,4) are the required triplets
, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)
{
Scanner sc = new Scanner(System.in);
int n = sc.nextInt();
int arr[] = new int[n];
for(int i = 0; i < n; i++)
{
arr[i] = sc.nextInt();
}
int x = sc.nextInt();
int cnt = 0, sum = 0;
for(int i=0;i<n-2;i++)
{
if(arr[i]>x)
{break;}
int tar=x-arr[i];
int j=i+1,k=n-1;
while(j<k){
if((arr[j]+arr[k])<tar){j++;}
else if((arr[j]+arr[k])>tar){k--;}
else{cnt++;j++;}
}
}
System.out.println(cnt);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Bruce Banner is working on a code involving 3 numbers with their sum as a mystery number. Just then the emergency alarm is sounded and Bruce hulks his way out to save the city of New York yet again. The responsibility of cracking the code is then handed over to you since it is crucial to ensure Hulk’s safety. You find the mystery number as X. Now give Hulk a number of all the triplets and help Hulk as he fights the evil Chitauri.First line of input contains a single integer N (length of the array)
second line contain the array elements
third line contain a single integer K(sum)
Constraint:-
1<=N<=1000
1<=elements<=100000
1<=K<=100000Output a single line containing the number of required tripletsSample Input:
6
1 2 3 4 5 6
8
Sample Output:-
2
Explanation:-
(1,2,5) , (1,3,4) are the required triplets
, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
int n;
cin>>n;
int a[n];
for(int i=0;i<n;i++){
cin>>a[i];
}
int x;cin>>x;
long cnt=0;
long sum=0;
for(int i=0;i<n-2;i++){
if(a[i]>x){break;}
int tar=x-a[i];
int j=i+1,k=n-1;
while(j<k){
if((a[j]+a[k])<tar){j++;}
else if((a[j]+a[k])>tar){k--;}
else{cnt++;j++;}
}
}
cout<<cnt<<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 of size n consisting of unique integers and an integer k. A function f is defined as follows:
f(i) = Minimum j such that subarray a<sub>i</sub>, a<sub>i+1</sub>, ., a<sub>j</sub> have k elements strictly greater than a<sub>i</sub>
and -1 (if no such j exists)
Your task is to calculate f(i) for all 1 ≤ i ≤ n.The first line contains an integer t (1 <= T <= 10<sup>5</sup>) - the number of test cases.
The first line of each test case contains two integers n and k (1 <= n <= 10<sup>5</sup>, 1 <= k <= n).
The second line of each test case contains n integers a<sub>1</sub>, a<sub>2</sub>, …a<sub>n</sub> (−10<sup>9</sup> <= a<sub>i</sub> <= 10<sup>9</sup>).
It is guaranteed that elements in a given array are unique.
It is guaranteed that the sum of n over all test cases does not exceed 2 * 10<sup>5</sup>.For each test case, print a separate line containing space-separated n integers f(1), f(2),. , f(n).Sample Input 1:
1
3 1
2 1 3
Sample Output 1:
2 2 -1
Explanation:
For index 0, the element at index 2 is the first element greater than 2.
For index 1, the element at index 2 is the first element greater than 1.
For index 2, no such index exists.
, I have written this Solution Code: import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main {
public static void process() throws IOException {
int n = sc.nextInt(),k = sc.nextInt();
FenwickTree tree = new FenwickTree(n+1);
int arr[] = sc.readArray(n);
compress(arr);
ArrayList<Pair> q = new ArrayList<Main.Pair>();
for(int i = 1; i<=n; i++)q.add(new Pair(arr[i-1], i));
Collections.sort(q);
int[] res = new int[n];
for(int i = 1; i<=n; i++) {
Pair curr = q.get(i-1);
int l = curr.y,r = n;
int ans = -1;
while(l<=r) {
int mid = (l+r)/2;
int len = (mid-curr.y+1);
len-=tree.find(curr.y,mid);
if(len > k) {
ans = mid-1;
r = mid-1;
}
else {
l = mid+1;
}
}
res[curr.y-1] = ans;
tree.add(curr.y, 1);
}
for(int e : res)out.print(e+" ");
out.println();
}
static class FenwickTree
{
public int[] tree;
public int size;
public FenwickTree(int size)
{
this.size = size;
tree = new int[size+5];
}
public void add(int i, int v)
{
while(i <= size)
{
tree[i] += v;
i += i&-i;
}
}
public int find(int i)
{
int res = 0;
while(i >= 1)
{
res += tree[i];
i -= i&-i;
}
return res;
}
public int find(int l, int r)
{
return find(r)-find(l-1);
}
}
private static long INF = 2000000000000000000L, M = 1000000007, MM = 998244353;
private static int N = 2_000_00;
private static void google(int tt) {
System.out.print("Case #" + (tt) + ": ");
}
static FastScanner sc;
static FastWriter out;
public static void main(String[] args) throws IOException {
boolean oj = true;
if (oj) {
sc = new FastScanner();
out = new FastWriter(System.out);
} else {
sc = new FastScanner("input.txt");
out = new FastWriter("output.txt");
}
long s = System.currentTimeMillis();
int t = 1;
t = sc.nextInt();
int TTT = 1;
while (t-- > 0) {
process();
}
out.flush();
}
private static boolean oj = System.getProperty("ONLINE_JUDGE") != null;
private static void tr(Object... o) {
if (!oj)
System.err.println(Arrays.deepToString(o));
}
static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
@Override
public int compareTo(Pair o) {
return Integer.compare(this.x, o.x);
}
}
static int ceil(int x, int y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long ceil(long x, long y) {
return (x % y == 0 ? x / y : (x / y + 1));
}
static long sqrt(long z) {
long sqz = (long) Math.sqrt(z);
while (sqz * 1L * sqz < z) {
sqz++;
}
while (sqz * 1L * sqz > z) {
sqz--;
}
return sqz;
}
static int log2(int N) {
int result = (int) (Math.log(N) / Math.log(2));
return result;
}
public static long gcd(long a, long b) {
if (a > b)
a = (a + b) - (b = a);
if (a == 0L)
return b;
return gcd(b % a, a);
}
public static long lcm(long a, long b) {
return (a * b) / gcd(a, b);
}
public static int lower_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = -1;
while (low <= high) {
mid = (low + high) / 2;
if (arr[mid] > x) {
high = mid - 1;
} else {
ans = mid;
low = mid + 1;
}
}
return ans;
}
public static int upper_bound(int[] arr, int x) {
int low = 0, high = arr.length - 1, mid = -1;
int ans = arr.length;
while (low < high) {
mid = (low + high) / 2;
if (arr[mid] >= x) {
ans = mid;
high = mid - 1;
} else {
low = mid + 1;
}
}
return ans;
}
static void ruffleSort(int[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
int temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void ruffleSort(long[] a) {
Random get = new Random();
for (int i = 0; i < a.length; i++) {
int r = get.nextInt(a.length);
long temp = a[i];
a[i] = a[r];
a[r] = temp;
}
Arrays.sort(a);
}
static void reverseArray(int[] a) {
int n = a.length;
int arr[] = new int[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
static void reverseArray(long[] a) {
int n = a.length;
long arr[] = new long[n];
for (int i = 0; i < n; i++)
arr[i] = a[n - i - 1];
for (int i = 0; i < n; i++)
a[i] = arr[i];
}
public static void push(TreeMap<Integer, Integer> map, int k, int v) {
if (!map.containsKey(k))
map.put(k, v);
else
map.put(k, map.get(k) + v);
}
public static void pull(TreeMap<Integer, Integer> map, int k, int v) {
int lol = map.get(k);
if (lol == v)
map.remove(k);
else
map.put(k, lol - v);
}
public static int[] compress(int[] arr) {
ArrayList<Integer> ls = new ArrayList<Integer>();
for (int x : arr)
ls.add(x);
Collections.sort(ls);
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int boof = 1;
for (int x : ls)
if (!map.containsKey(x))
map.put(x, boof++);
int[] brr = new int[arr.length];
for (int i = 0; i < arr.length; i++)
brr[i] = map.get(arr[i]);
return brr;
}
public static class FastWriter {
private static final int BUF_SIZE = 1 << 13;
private final byte[] buf = new byte[BUF_SIZE];
private final OutputStream out;
private int ptr = 0;
private FastWriter() {
out = null;
}
public FastWriter(OutputStream os) {
this.out = os;
}
public FastWriter(String path) {
try {
this.out = new FileOutputStream(path);
} catch (FileNotFoundException e) {
throw new RuntimeException("FastWriter");
}
}
public FastWriter write(byte b) {
buf[ptr++] = b;
if (ptr == BUF_SIZE)
innerflush();
return this;
}
public FastWriter write(char c) {
return write((byte) c);
}
public FastWriter write(char[] s) {
for (char c : s) {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
}
return this;
}
public FastWriter write(String s) {
s.chars().forEach(c -> {
buf[ptr++] = (byte) c;
if (ptr == BUF_SIZE)
innerflush();
});
return this;
}
private static int countDigits(int l) {
if (l >= 1000000000)
return 10;
if (l >= 100000000)
return 9;
if (l >= 10000000)
return 8;
if (l >= 1000000)
return 7;
if (l >= 100000)
return 6;
if (l >= 10000)
return 5;
if (l >= 1000)
return 4;
if (l >= 100)
return 3;
if (l >= 10)
return 2;
return 1;
}
public FastWriter write(int x) {
if (x == Integer.MIN_VALUE) {
return write((long) x);
}
if (ptr + 12 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
private static int countDigits(long l) {
if (l >= 1000000000000000000L)
return 19;
if (l >= 100000000000000000L)
return 18;
if (l >= 10000000000000000L)
return 17;
if (l >= 1000000000000000L)
return 16;
if (l >= 100000000000000L)
return 15;
if (l >= 10000000000000L)
return 14;
if (l >= 1000000000000L)
return 13;
if (l >= 100000000000L)
return 12;
if (l >= 10000000000L)
return 11;
if (l >= 1000000000L)
return 10;
if (l >= 100000000L)
return 9;
if (l >= 10000000L)
return 8;
if (l >= 1000000L)
return 7;
if (l >= 100000L)
return 6;
if (l >= 10000L)
return 5;
if (l >= 1000L)
return 4;
if (l >= 100L)
return 3;
if (l >= 10L)
return 2;
return 1;
}
public FastWriter write(long x) {
if (x == Long.MIN_VALUE) {
return write("" + x);
}
if (ptr + 21 >= BUF_SIZE)
innerflush();
if (x < 0) {
write((byte) '-');
x = -x;
}
int d = countDigits(x);
for (int i = ptr + d - 1; i >= ptr; i--) {
buf[i] = (byte) ('0' + x % 10);
x /= 10;
}
ptr += d;
return this;
}
public FastWriter write(double x, int precision) {
if (x < 0) {
write('-');
x = -x;
}
x += Math.pow(10, -precision) / 2;
write((long) x).write(".");
x -= (long) x;
for (int i = 0; i < precision; i++) {
x *= 10;
write((char) ('0' + (int) x));
x -= (int) x;
}
return this;
}
public FastWriter writeln(char c) {
return write(c).writeln();
}
public FastWriter writeln(int x) {
return write(x).writeln();
}
public FastWriter writeln(long x) {
return write(x).writeln();
}
public FastWriter writeln(double x, int precision) {
return write(x, precision).writeln();
}
public FastWriter write(int... xs) {
boolean first = true;
for (int x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter write(long... xs) {
boolean first = true;
for (long x : xs) {
if (!first)
write(' ');
first = false;
write(x);
}
return this;
}
public FastWriter writeln() {
return write((byte) '\n');
}
public FastWriter writeln(int... xs) {
return write(xs).writeln();
}
public FastWriter writeln(long... xs) {
return write(xs).writeln();
}
public FastWriter writeln(char[] line) {
return write(line).writeln();
}
public FastWriter writeln(char[]... map) {
for (char[] line : map)
write(line).writeln();
return this;
}
public FastWriter writeln(String s) {
return write(s).writeln();
}
private void innerflush() {
try {
out.write(buf, 0, ptr);
ptr = 0;
} catch (IOException e) {
throw new RuntimeException("innerflush");
}
}
public void flush() {
innerflush();
try {
out.flush();
} catch (IOException e) {
throw new RuntimeException("flush");
}
}
public FastWriter print(byte b) {
return write(b);
}
public FastWriter print(char c) {
return write(c);
}
public FastWriter print(char[] s) {
return write(s);
}
public FastWriter print(String s) {
return write(s);
}
public FastWriter print(int x) {
return write(x);
}
public FastWriter print(long x) {
return write(x);
}
public FastWriter print(double x, int precision) {
return write(x, precision);
}
public FastWriter println(char c) {
return writeln(c);
}
public FastWriter println(int x) {
return writeln(x);
}
public FastWriter println(long x) {
return writeln(x);
}
public FastWriter println(double x, int precision) {
return writeln(x, precision);
}
public FastWriter print(int... xs) {
return write(xs);
}
public FastWriter print(long... xs) {
return write(xs);
}
public FastWriter println(int... xs) {
return writeln(xs);
}
public FastWriter println(long... xs) {
return writeln(xs);
}
public FastWriter println(char[] line) {
return writeln(line);
}
public FastWriter println(char[]... map) {
return writeln(map);
}
public FastWriter println(String s) {
return writeln(s);
}
public FastWriter println() {
return writeln();
}
}
static class FastScanner {
private int BS = 1 << 16;
private char NC = (char) 0;
private byte[] buf = new byte[BS];
private int bId = 0, size = 0;
private char c = NC;
private double cnt = 1;
private BufferedInputStream in;
public FastScanner() {
in = new BufferedInputStream(System.in, BS);
}
public FastScanner(String s) {
try {
in = new BufferedInputStream(new FileInputStream(new File(s)), BS);
} catch (Exception e) {
in = new BufferedInputStream(System.in, BS);
}
}
private char getChar() {
while (bId == size) {
try {
size = in.read(buf);
} catch (Exception e) {
return NC;
}
if (size == -1)
return NC;
bId = 0;
}
return (char) buf[bId++];
}
public int nextInt() {
return (int) nextLong();
}
public int[] readArray(int N) {
int[] res = new int[N];
for (int i = 0; i < N; i++) {
res[i] = (int) nextLong();
}
return res;
}
public long[] readArrayLong(int N) {
long[] res = new long[N];
for (int i = 0; i < N; i++) {
res[i] = nextLong();
}
return res;
}
public int[][] readArrayMatrix(int N, int M, int Index) {
if (Index == 0) {
int[][] res = new int[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
int[][] res = new int[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = (int) nextLong();
}
return res;
}
public long[][] readArrayMatrixLong(int N, int M, int Index) {
if (Index == 0) {
long[][] res = new long[N][M];
for (int i = 0; i < N; i++) {
for (int j = 0; j < M; j++)
res[i][j] = nextLong();
}
return res;
}
long[][] res = new long[N][M];
for (int i = 1; i <= N; i++) {
for (int j = 1; j <= M; j++)
res[i][j] = nextLong();
}
return res;
}
public long nextLong() {
cnt = 1;
boolean neg = false;
if (c == NC)
c = getChar();
for (; (c < '0' || c > '9'); c = getChar()) {
if (c == '-')
neg = true;
}
long res = 0;
for (; c >= '0' && c <= '9'; c = getChar()) {
res = (res << 3) + (res << 1) + c - '0';
cnt *= 10;
}
return neg ? -res : res;
}
public double nextDouble() {
double cur = nextLong();
return c != '.' ? cur : cur + nextLong() / cnt;
}
public double[] readArrayDouble(int N) {
double[] res = new double[N];
for (int i = 0; i < N; i++) {
res[i] = nextDouble();
}
return res;
}
public String next() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c > 32) {
res.append(c);
c = getChar();
}
return res.toString();
}
public String nextLine() {
StringBuilder res = new StringBuilder();
while (c <= 32)
c = getChar();
while (c != '\n') {
res.append(c);
c = getChar();
}
return res.toString();
}
public boolean hasNext() {
if (c > 32)
return true;
while (true) {
c = getChar();
if (c == NC)
return false;
else if (c > 32)
return true;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array of size n consisting of unique integers and an integer k. A function f is defined as follows:
f(i) = Minimum j such that subarray a<sub>i</sub>, a<sub>i+1</sub>, ., a<sub>j</sub> have k elements strictly greater than a<sub>i</sub>
and -1 (if no such j exists)
Your task is to calculate f(i) for all 1 ≤ i ≤ n.The first line contains an integer t (1 <= T <= 10<sup>5</sup>) - the number of test cases.
The first line of each test case contains two integers n and k (1 <= n <= 10<sup>5</sup>, 1 <= k <= n).
The second line of each test case contains n integers a<sub>1</sub>, a<sub>2</sub>, …a<sub>n</sub> (−10<sup>9</sup> <= a<sub>i</sub> <= 10<sup>9</sup>).
It is guaranteed that elements in a given array are unique.
It is guaranteed that the sum of n over all test cases does not exceed 2 * 10<sup>5</sup>.For each test case, print a separate line containing space-separated n integers f(1), f(2),. , f(n).Sample Input 1:
1
3 1
2 1 3
Sample Output 1:
2 2 -1
Explanation:
For index 0, the element at index 2 is the first element greater than 2.
For index 1, the element at index 2 is the first element greater than 1.
For index 2, no such index exists.
, I have written this Solution Code: #include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.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>;
#define endl '\n'
#define pb push_back
#define ub upper_bound
#define lb lower_bound
#define fi first
#define se second
#define int long long
typedef long long ll;
typedef long double ld;
#define pii pair<int,int>
#define sz(x) ((ll)x.size())
#define fr(a,b,c) for(int a=b; a<=c; a++)
#define frev(a,b,c) for(int a=c; a>=b; a--)
#define rep(a,b,c) for(int a=b; a<c; a++)
#define trav(a,x) for(auto &a:x)
#define all(con) con.begin(),con.end()
#define done(x) {cout << x << endl;return;}
#define mini(x,y) x = min(x,y)
#define maxi(x,y) x = max(x,y)
const ll infl = 0x3f3f3f3f3f3f3f3fLL;
const int infi = 0x3f3f3f3f;
mt19937_64 mt(chrono::steady_clock::now().time_since_epoch().count());
//const int mod = 998244353;
const int mod = 1e9 + 7;
typedef vector<int> vi;
typedef vector<string> vs;
typedef vector<vector<int>> vvi;
typedef vector<pair<int, int>> vpii;
typedef map<int, int> mii;
typedef set<int> si;
typedef set<pair<int,int>> spii;
typedef queue<int> qi;
uniform_int_distribution<int> rng(0, 1e9);
// DEBUG FUNCTIONS START
void __print(int x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}
template<typename T, typename V> void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T> void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void deb() {cerr << "\n";}
template <typename T, typename... V> void deb(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; deb(v...);}
// DEBUG FUNCTIONS END
const int N = 2e5 + 5;
void solve(){
int n, k;
cin >> n >> k;
vi a(n);
rep(i,0,n){
cin >> a[i];
}
vi b = a;
sort(all(b));
vi pos(n);
ordered_set<int> s;
rep(i,0,n){
a[i] = lb(all(b), a[i]) - b.begin();
pos[a[i]] = i;
s.insert(i);
}
vi res(n, -1);
rep(i,0,n){
int position = pos[i];
int idx = s.order_of_key(position);
if(sz(s) > idx + k){
res[position] = *s.find_by_order(idx + k);
}
s.erase(position);
}
rep(i,0,n){
cout << res[i] << " ";
}
cout << endl;
}
signed main(){
ios_base::sync_with_stdio(0), cin.tie(0);
cout << fixed << setprecision(15);
//freopen ("03.txt","r",stdin);
//freopen ("3.txt","w",stdout);
int t = 1;
cin >> t;
while (t--)
solve();
return 0;
}
int powm(int a, int b){
int res = 1;
while (b) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are A Bacterias.
Each time Jerry shouts, the bacterias multiply by K times.
In order to have B or more slimes, at least how many times does Jerry need to shout?Input is given from Standard Input in the following format:
A B K
<b>Constraints</b>
1≤A≤B≤10^9
2≤K≤10^9
All values in input are integers.Print the answer.<b>Sample Input 1</b>
1 4 2
<b>Sample Output 1</b>
2
<b>Sample Input 2</b>
7 7 10
<b>Sample Output 2</b>
0
<b>Sample Input 3</b>
31 415926 5
<b>Sample Output 3</b>
6
(In first case, after the first shout Jerry will have 2 bacteria and after the second shout, he will have 4(2*2) bacteria., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define trav(a, x) for(auto& a : x)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
#define subnb true
#define Lnb true
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;
int main() {
ios::sync_with_stdio(false);
cin.tie(NULL);
ll a, b, k;
cin >> a >> b >> k;
int res = 0;
while(a < b) {
a *= k;
res++;
}
cout << res << '\n';
}, 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: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<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 <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
def checkSpecial_M_Visor(N,M) :
# Final result of summation of divisors
i=2;
count=0
while i<=N :
if(N%i==0):
count=count+1
i=i+2
if(count == M):
return "Yes"
return "No"
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<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 <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
static String checkSpecial_M_Visor(int N, int M)
{
int temp=0;
for(int i = 2; i <=N; i+=2)
{
if(N%i==0)
temp++;
}
if(temp==M)
return "Yes";
else return "No";
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<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 <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code:
string checkSpecial_M_Visor(int N,int M){
int count=0;
for(int i=2;i<=N;i+=2){
if(N%i==0){
count++;
}
}
if(count==M){return "Yes";}
return "No";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two positive integers N and M. The task is to check whether N is a <b>Special-M-visor</b> or not.
<b>Special-M-visor</b>: A number is called Special-m-visor if it has exactly M even divisors for a given N.<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 <b>checkSpecial_M_Visor()</b>. Where you will get N and M as a parameter.
Constraints:
1 <= N <= 10^6
0 <= M <= 10^6Return "Yes" without quotes if the number N is a <b>Special- M- visor</b> else return "No"Sample Input:-
4 2
Sample Output:-
Yes
Explanation:- 2 and 4 are the only even divisors of 4
Sample Input:-
8 5
Sample Output:-
No, I have written this Solution Code: const char* checkSpecial_M_Visor(int N,int M){
int count=0;
for(int i=2;i<=N;i+=2){
if(N%i==0){
count++;
}
}
char* ans;
if(count==M){return "Yes";}
return "No";
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, 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 in = br.readLine().trim();
boolean printed = false;
for(int i = 1; i <= (in.length() / 2); i++) {
ArrayList<Long> valueArray = new ArrayList<>();
boolean incremented = false;
int len = i;
for(int j = 0; j < in.length(); ) {
String temp = in.substring(j, Math.min((len+j), in.length()));
valueArray.add(Long.parseLong(temp));
j+=len;
if(Long.parseLong(temp) % 10 == 9) {
int digits = findLength(Long.parseLong(temp));
if(Long.parseLong(temp) % ((long)Math.pow(10, digits) - 1) == 0)
++len;
incremented = true;
}
}
long[] arrli = new long[valueArray.size()];
for(int k = 0; k < valueArray.size(); k++) {
arrli[k] = valueArray.get(k);
}
if(incremented) {
--len;
}
if(isBeauty(arrli)) {
System.out.println("YES " + arrli[0]);
printed = true;
break;
}
}
if(!printed)
System.out.println("NO");
}
public static int findLength(long n) {
int count = 0;
while(n > 0) {
n /= 10;
++count;
}
return count;
}
public static boolean isBeauty(long[] tempOne) {
for(int i = 1; i < tempOne.length; i++) {
if(tempOne[i-1] + 1 != tempOne[i])
return false;
}
return true;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, I have written this Solution Code: def isSequence(string):
start = 0
n = len(string)
for i in range(0,n//2):
new_str = string[0:i+1]
num = int(new_str)
start = num
while len(new_str) < n:
num +=1
new_str += str(num)
if(new_str == string):
return start
return -1
string = input()
start = isSequence(string)
if start != -1:
print("YES",end=" ")
print(start)
else:
print("NO"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A string is called beautiful if it can be represented by two or more consecutive numbers. Given a string S check if it is beautiful or not.
Note:- The numbers should not contain leading zeroes in it i. e string "10203" ("1" + "02" + "03") will not be treated as a beautiful stringInput contains a single string containing the string S.
Constraints:-
1 <= |S| <= 32Print "YES" if the given string is beautiful and print the first number of the sequence else print "NO".Sample Input:-
101112
Sample Output:-
YES 10
Explanation:-
Given string can be broken as "10" + "11" + "12"
Sample Input:-
101111
Sample Output:-
NO, I have written this Solution Code: #include <sstream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <string>
#include <cassert>
#include <ctime>
#include <map>
#include <math.h>
#include <cstdio>
#include <set>
#include <deque>
#include <memory.h>
#include <queue>
#pragma comment(linker, "/STACK:64000000")
typedef long long ll;
using namespace std;
const int MAXK = -1;
const int MAXN = -1;
const int MOD = 0; // 1000 * 1000 * 1000 + 7;
int main() {
#ifdef _MSC_VER
freopen("input.txt", "r", stdin);
#endif
int T;
T=1;
while (T--) {
string s;
cin >> s;
int n = s.length();
ll ans = -1;
for (int len = 1; len * 2 <= n; len++) {
ll x = 0;
for (int i = 0; i < len; i++) x = 10 * x + (s[i] - '0');
string t = "";
ll x0 = x;
while (t.length() < s.length()) t += to_string(x++);
if (s == t) {
ans = x0;
break;
}
}
if (ans == -1) cout << "NO" << endl;
else cout << "YES " << ans << endl;
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
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));
int size = Integer.parseInt(br.readLine());
String line[] = br.readLine().split(" ");
long a[] = new long[size + 1];
for(int i = 1 ; i <= size; i++)
a[i] = Long.parseLong(line[i - 1]);
long value1 = maxofsubarray(a,size);
long value2 = maxofsubarrayBreakingatNegativeVal(a, size);
System.out.print(Math.max(value1, value2));
}
static long maxofsubarray(long[] a , int size)
{
long currsum = 0 , maxsum = 0, maxEl = 0;
for(int i = 1 ; i <= size ; i++)
{
currsum += a[i];
if(currsum < 0) currsum =0;
if(currsum > maxsum)
{
maxsum = currsum; maxEl = Math.max(maxEl, a[i]);
}
} return (maxsum - maxEl);
}
static long maxofsubarrayBreakingatNegativeVal(long[] a , int size)
{
long currsum = 0 , maxsum = 0, maxEl = 0, maxofall = 0;
for(int i = 1 ; i <= size ; i++)
{
maxEl = Math.max(maxEl, a[i]);
currsum += a[i];
if(currsum - maxEl < 0)
{
currsum =0; maxEl = 0;
}
if(currsum - maxEl > maxsum)
{
maxofall = currsum - maxEl; maxsum = maxofall;
}
}
return (maxofall);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
0, I have written this Solution Code: def maxSumTill(arr,n):
currSum = 0
maxSum = 0
maxEle = 0
maxi = 0
for i in range(1,n+1):
maxEle = max(maxEle,arr[i])
currSum += arr[i]
if currSum-maxEle < 0:
currSum = 0
maxEle =0
if currSum-maxEle > maxSum:
maxi = currSum - maxEle
maxSum = maxi
return maxi
def maxOfSum(arr,n):
currSum = 0
maxSum = 0
maxEle = 0
for i in range(1,n+1):
currSum += arr[i]
if currSum < 0:
currSum = 0
if currSum > maxSum:
maxSum = currSum
maxEle = max(maxEle, arr[i])
return maxSum - maxEle
n = int(input())
arr = list(map(int,input().split()))
arr = [0]+arr
val1 = maxSumTill(arr,n)
val2 = maxOfSum(arr,n)
print(max(val1,val2))
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Tono and Solo are one of the best programmers on the planet. Solo knows how to solve the problem on maximum subarray sum using the Kadane algorithm (don't you?). Now, Tono gives Solo a trickier problem to solve. In this problem, Solo can choose any non- empty subarray, the value of the subarray will be the <b>(sum of all the elements of the subarray - largest element of the subarray)</b>. Can you find the maximum value of any subarray?The first line of the input contains a single integer N, the size of the array.
The second line of the input contains N elements A[1], A[2],. , A[N].
Constraints
1 <= N <= 100000
-30 <= A[i] <= 30Output a single integer, the value of the subarray having maximum value.Sample Input
7
5 2 5 3 -30 -30 6
Sample Output
10
Explanation: Solo chooses the subarray [5, 2, 5, 3]. The value of the subarray is 10.
Sample Input
3
-10 6 -15
Sample Output
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(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;
const int N = 1e5+5;
// 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
int arr[N];
int n;
int kadane(){
int mx = 0;
int cur = 0;
For(i, 0, n){
cur += arr[i];
mx = max(mx, cur);
if(cur < 0)
cur = 0;
}
return mx;
}
void play(int x){
For(i, 0, n){
if(arr[i]==x){
arr[i]=-1000000000;
}
}
}
void solve(){
cin>>n;
For(i, 0, n){
cin>>arr[i];
assert(arr[i]>=-30 && arr[i]<=30);
}
int ans = 0;
for(int i=30; i>=1; i--){
ans = max(ans, kadane()-i);
play(i);
}
cout<<ans;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
int t=1;
// cin>>t;
while(t--){
solve();
cout<<"\n";
}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, I have written this Solution Code: for _ in range(int(input())):
n = int(input())
if n%2 or n<6:
print(-1)
continue
m = n//2
sl = 1
while m % 2 == 0:
m >>= 1
sl <<= 1
if n == sl*2:
print(-1)
continue
print(sl,n//2-sl,n//2), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, 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>;
// PRAGMAS (do these even work?)
#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,avx,avx2")
#pragma GCC optimization ("O3")
#pragma GCC optimization ("unroll-loops")
// 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 long double ld;
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);
}
};
int main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
ll t;
cin >> t;
REP(i, 0, t)
{
ll n;
cin >> n;
if(n&1) cout << "-1\n";
else
{
ll x = n/2;
vll bits;
REP(i, 0, 60)
{
if((x >> i)&1)
{
bits.pb(i);
}
}
if(bits.size() == 1)
{
cout << "-1\n";
}
else
{
ll a = (1ll << bits[0]);
cout << a << " " << x - a << " " << x << "\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: For a given positive integer N, determine if there exist three <b>positive</b> integers a, b and c such that the following two conditions hold:
1. a + b + c = N
2. a ^ b ^ c = 0 where ^ denotes the bitwise XOR operation.
If there exist such a triple (a, b, c), print the lexicographically smallest one.
Else, print -1.The first line of input contains a single integer, T.
T lines follow, each containing a single integer, N.
<b>Constraints:</b>
1 <= T <= 10<sup>3</sup>
3 <= N <= 10<sup>18</sup>For each test, in a new line, print the lexicographically smallest triple (a, b, c) if it exists, else print -1.Sample Input:
3
3
6
12
Sample Output:
-1
1 2 3
2 4 6, I have written this Solution Code:
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.*;
import static java.lang.Math.*;
import static java.lang.System.out;
public class Main {
void solve(){
long n= in.nextLong();
if(n%2==1){
sb.append(-1).append("\n");
}
else{
long cnt=0;
n=n/2;
while(n%2==0){
cnt++;
n=n>>1;
}
long x=(1L<<cnt);
long y=(1L<<cnt);
long z=0;
while(n!=0){
n=n>>1;
cnt++;
if((n&1)==1){
y+=(1L<<cnt);
z+=(1L<<cnt);
}
}
long[] a= new long[3];
a[0]=x;
a[1]=y;
a[2]=z;
sort(a);
if(a[0]!=0){
sb.append(a[0]).append(" ");
sb.append(a[1]).append(" ");
sb.append(a[2]).append("\n");
}
else{
sb.append(-1).append("\n");
}
}
}
FastReader in;
StringBuffer sb;
public static void main(String[] args) {
new Main().run();
}
void run(){
in= new FastReader();
start();
}
void start(){
sb= new StringBuffer();
for(int t=in.nextInt();t>0;t--) {
solve();
}
out.print(sb);
}
void swap( int i , int j) {
int tmp = i;
i = j;
j = tmp;
}
long power(long x, long y, long p) {
long res = 1;
x = x % p;
while (y > 0) {
if (y % 2 == 1)
res = (res * x) % p;
y = y >> 1;
x = (x * x) % p;
}
return res;
}
int lower_bound(long[] a, long x) {
int l=-1,r=a.length;
while(l+1<r) {
int m=(l+r)>>>1;
if(a[m]>=x) r=m;
else l=m;
}
return r;
}
int upper_bound(long[] arr, int key) {
int i=0, j=arr.length-1;
if (arr[j]<=key) return j+1;
if(arr[i]>key) return i;
while (i<j){
int mid= (i+j)/2;
if(arr[mid]<=key){
i= mid+1;
}else{
j=mid;
}
}
return i;
}
void sort(long[] A){
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i){
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
int[] intArr(int n){
int[] res= new int[n];
for(int i=0;i<n;i++){
res[i]= in.nextInt();
}
return res;
}
long[] longArr(int n){
long[] res= new long[n];
for(int i=0;i<n;i++){
res[i]= in.nextLong();
}
return res;
}
boolean isDigitSumPalindrome(long N) {
long sum= sumOfDigits(String.valueOf(N));
long rev=0;
long org= sum;
while (sum!=0){
long d= sum%10;
rev = rev*10 +d;
sum /= 10;
}
return org == rev;
}
long sumOfDigits(String n){
long sum= 0;
for (char c: n.toCharArray()){
sum += Integer.parseInt(String.valueOf(c));
}
return sum;
}
long[] revArray(long[] arr) {
int n= arr.length;
int i=0, j=n-1;
while (i<j){
long temp= arr[i];
arr[i]= arr[j];
arr[j]= temp;
i++;
j--;
}
return arr;
}
long gcd(long a, long b){
if (b==0)
return a;
return gcd(b, a%b);
}
long lcm(long a,long b){
return (a*b)/gcd(a,b);
}
static class Pair implements Comparable<Pair>{
long first;
long second;
Pair(long x, long y){
this.first=x;
this.second=y;
}
@Override
public int compareTo(Pair o) {
return 0;
}
}
public 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 (Exception e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
float nextFloat(){
return Float.parseFloat(next());
}
String nextLine(){
String str="";
try{
str=br.readLine();
}catch (Exception e){
e.printStackTrace();
}
return str;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A tuple (R, G, B) is considered good if R*G+B <= N and R, G and B are all positive. Given N, find number of good tuples.Input contains a single integer N.
Constraints:
1 <= N <= 1000000Print number of good tuples.Sample Input
3
Sample Output
4
Explanation:
Following are the good tuples:
(1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(2, 1, 1), I have written this Solution Code: import java.util.*;
import java.io.*;
class Main{
public static void main(String[] args)throws IOException{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int n=Integer.parseInt(br.readLine());
long count=0;
for(int i=1;i<n;i++){
int t=n-i;
while(t>=1){
count+=t;
t=t-i;
}
}
System.out.println(count);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A tuple (R, G, B) is considered good if R*G+B <= N and R, G and B are all positive. Given N, find number of good tuples.Input contains a single integer N.
Constraints:
1 <= N <= 1000000Print number of good tuples.Sample Input
3
Sample Output
4
Explanation:
Following are the good tuples:
(1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(2, 1, 1), I have written this Solution Code: n = int(input())
m = n-1
count = 0
for r in range(1, n):
c = m//r
count += c*(n-r*c) + r*c*(c-1)//2
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: A tuple (R, G, B) is considered good if R*G+B <= N and R, G and B are all positive. Given N, find number of good tuples.Input contains a single integer N.
Constraints:
1 <= N <= 1000000Print number of good tuples.Sample Input
3
Sample Output
4
Explanation:
Following are the good tuples:
(1, 1, 1)
(1, 1, 2)
(1, 2, 1)
(2, 1, 1), 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 ans=0;
for(int i=1;i<=n;++i)
{
int j=i;
while(j<=n){
ans+=n-j;
j+=i;
}
}
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: Write a recursive program to remove all tabs or spaces from a string.The first line contains the string s.
Constraints:
0<len(s)<=100Prints the string after removing all the tabs and spaces.Sample Input:
Hello World
Sample Output:
HelloWorld
Explanation:
The string "Hello World" after removing spaces is "HelloWorld"., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
int n=0;
BufferedReader br =new BufferedReader(new InputStreamReader(System.in));
String inputString=br.readLine();
removeSpaces(inputString,n);
}
catch(Exception e){
}
}
public static void removeSpaces(String inputString,int n){
if(n>inputString.length()){
return;
}
if(n<inputString.length()&& ((inputString.charAt(n)!=' ') ))
{
System.out.print(inputString.charAt(n));
removeSpaces(inputString,n+1);
}
else if(n<inputString.length()){
removeSpaces(inputString,n+1);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a recursive program to remove all tabs or spaces from a string.The first line contains the string s.
Constraints:
0<len(s)<=100Prints the string after removing all the tabs and spaces.Sample Input:
Hello World
Sample Output:
HelloWorld
Explanation:
The string "Hello World" after removing spaces is "HelloWorld"., I have written this Solution Code: def remove(string):
# Base Case
if not string:
return ""
# Recursive Case
if string[0] == "\t" or string[0] == " ":
return remove(string[1:])
else:
return string[0] + remove(string[1:])
# Driver Code
s=input()
print(remove(s)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given 2 numbers a, b.
You can perform at most b steps
In one step :
1- increase a by 3
2- decrease a by 3
3- multiply a by 2
Find number of distinct numbers we can make after performing at most b operations on a.The first line contains the number of tests T.
For each test case:
Input two integers a and b.
0 < T <= 100
1 <= a <= 100000
1 <= b <= 9Print answer on a separate line for each test caseSample Input
2
5 1
30 2
Sample Output
4
11
For test 1:-
In 0 steps, 5 can be formed
In 1 steps 2, 8, 10 can be formed
For test 2:-
in 0 step :- 30
in 1 step- 27 33 60
in 2 step:- 24, 30, 54, 30, 36, 66, 57 63 120
total unique number = 11, I have written this Solution Code: import java.util.*;
class Main{
public static void main(String args[]){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int a=sc.nextInt();
int b=sc.nextInt();
ArrayList<Integer> arr=new ArrayList();
arr.add(a);
for(int j=1;j<b+1;j++)
{ int decrease=0;
int increse=0;
int mul=0;
int len=arr.size();
for(int e=0;e<len;e++)
{
decrease=arr.get(e)-3;
if(!arr.contains(decrease))
{
arr.add(decrease);
}
increse=arr.get(e)+3;
if(!arr.contains(increse))
{
arr.add(increse);
}
mul=arr.get(e)*2;
if(!arr.contains(mul))
{
arr.add(mul);
}
}
}
System.out.println(arr.size());
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given 2 numbers a, b.
You can perform at most b steps
In one step :
1- increase a by 3
2- decrease a by 3
3- multiply a by 2
Find number of distinct numbers we can make after performing at most b operations on a.The first line contains the number of tests T.
For each test case:
Input two integers a and b.
0 < T <= 100
1 <= a <= 100000
1 <= b <= 9Print answer on a separate line for each test caseSample Input
2
5 1
30 2
Sample Output
4
11
For test 1:-
In 0 steps, 5 can be formed
In 1 steps 2, 8, 10 can be formed
For test 2:-
in 0 step :- 30
in 1 step- 27 33 60
in 2 step:- 24, 30, 54, 30, 36, 66, 57 63 120
total unique number = 11, I have written this Solution Code: T = int(input())
for t in range(T):
n, k = map(int, input().strip().split())
s = set()
s.add(n)
l = list(s)
while(k):
for i in range(len(l)):
s.add(l[i]-3)
s.add(l[i]+3)
s.add(l[i]*2)
l = list(s)
k -= 1
print(len(s)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given 2 numbers a, b.
You can perform at most b steps
In one step :
1- increase a by 3
2- decrease a by 3
3- multiply a by 2
Find number of distinct numbers we can make after performing at most b operations on a.The first line contains the number of tests T.
For each test case:
Input two integers a and b.
0 < T <= 100
1 <= a <= 100000
1 <= b <= 9Print answer on a separate line for each test caseSample Input
2
5 1
30 2
Sample Output
4
11
For test 1:-
In 0 steps, 5 can be formed
In 1 steps 2, 8, 10 can be formed
For test 2:-
in 0 step :- 30
in 1 step- 27 33 60
in 2 step:- 24, 30, 54, 30, 36, 66, 57 63 120
total unique number = 11, 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
set<int> ss;
int yy;
void solve(int a,int x)
{
ss.insert(a);
if(yy<=x) return;
solve((a+3LL),x+1);
solve((a-3LL),x+1);
solve((a*2LL),x+1);
}
signed main()
{
int t;
cin>>t;
while(t>0)
{
t--;
int a,b;
cin>>a>>b;
ss.clear();
yy=b+1;
ss.insert(a);
solve(a,1);
cout<<ss.size()<<endl;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., 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 s1 = br.readLine();
String s2 = br.readLine();
boolean flag = true;
int[] arr1 = new int[26];
int[] arr2 = new int[26];
for(int i=0; i<s1.length(); i++){
arr1[s1.charAt(i)-97]++;
}
for(int i=0; i<s2.length(); i++){
arr2[s2.charAt(i)-97]++;
}
for(int i=0; i<25; i++){
if(arr1[i]!=arr2[i]){
flag = false;
break;
}
}
if(flag==true)
System.out.print("YES");
else System.out.print("NO");
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: s1 = input().strip()
s2 = input().strip()
dict1 = dict()
dict2 = dict()
for i in s1:
dict1[i] = dict1.get(i, 0) + 1
for j in s2:
dict2[j] = dict2.get(j, 0) + 1
print(("NO", "YES")[dict1 == dict2]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., 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
int A[26],B[26];
signed main()
{
string s,p;
cin>>s>>p;
for(int i=0;i<s.size();i++)
{
int y=s[i]-'a';
A[y]++;
}
for(int i=0;i<p.size();i++)
{
int y=p[i]-'a';
B[y]++;
}int ch=1;
for(int i=0;i<26;i++)
{
if(B[i]!=A[i])ch=0;
}
if(ch==1) cout<<"YES"<<endl;
else cout<<"NO"<<endl;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two strings a and b consisting of lowercase characters. The task is to check whether two given strings are an anagram of each other or not. An anagram of a string is another string that contains the same characters, only the order of characters can be different. For example, “act” and “tac” are an anagram of each other.Input consists of two strings in lowercase english characters.
Constraints:
1 ≤ |s1|, |s2| ≤ 10^5Print "YES" without quotes if the two strings are anagram else print "NO".Sample Input
naman
manan
Sample Output
YES
Explanation: Both String contain 2 'a's, 2 'n's and 1 'm'., I have written this Solution Code: // str1 and str2 are the two input strings
function isAnagram(str1,str2){
// Get lengths of both strings
let n1 = str1.length;
let n2 = str2.length;
// If length of both strings is not same,
// then they cannot be anagram
if (n1 != n2)
return "NO";
str1 = str1.split('')
str2 = str2.split('')
// Sort both strings
str1.sort();
str2.sort()
// Compare sorted strings
for (let i = 0; i < n1; i++)
if (str1[i] != str2[i])
return "NO";
return "YES";
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., 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;
int a[N];
signed main() {
IOS;
int t; cin >> t;
while(t--){
int n; cin >> n;
for(int i = 1; i <= n; i++)
cin >> a[i];
sort(a + 1, a + n + 1);
for(int i = 1; i <= n; i++)
cout << a[i] << " ";
cout << 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 A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: // arr is unsorted array
// n is the number of elements in the array
function bubbleSort(arr, n) {
// write code here
// do not console.log the answer
// return sorted array
return arr.sort((a, b) => a - b)
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., 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 t = sc.nextInt();
while(t-->0){
int n = sc.nextInt();
int a[] = new int[n];
for(int i=0;i<n;i++){
a[i]=sc.nextInt();
}
int temp;
for(int i=1;i<n;i++){
if(a[i]<a[i-1]){
for(int j=i;j>0;j--){
if(a[j]<a[j-1]){
temp=a[j];
a[j]=a[j-1];
a[j-1]=temp;
}
else{
break;
}
}
}
}
for(int i=0;i<n;i++){
System.out.print(a[i]+" ");
}
System.out.println();
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array A[], of size N containing positive integers. You need to print the elements of an array in increasing order.The first line of the input denotes the number of test cases 'T'. The first line of the test case is the size of the array and the second line consists of array elements.
For Python Users just complete the given function.
<b>Constraints:</b>
1 ≤ T ≤ 100
1 ≤ N ≤ 1000
1 ≤ A[i] ≤ 1000For each testcase print the sorted array in a new line.Input:
2
5
4 1 3 9 7
10
10 9 8 7 6 5 4 3 2 1
Output:
1 3 4 7 9
1 2 3 4 5 6 7 8 9 10
<b>Explanation:</b>
Testcase 1: 1 3 4 7 9 are in sorted form.
Testcase 2: For the given input, 1 2 3 4 5 6 7 8 9 10 are in sorted form., I have written this Solution Code: def bubbleSort(arr):
arr.sort()
return arr
, In this Programming Language: Python, 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: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
bool EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
int main(){
int n1,n2,v1,v2;
cin>>n1>>n2>>v1>>v2;
if(EqualOrNot(n1,n2,v1,v2)){
cout<<"Yes";}
else{
cout<<"No";
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: def EqualOrNot(h1,h2,v1,v2):
if (v2>v1 and (h1-h2)%(v2-v1)==0):
return True
else:
return False
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Gian and Suneo want their heights to be equal so they asked Doraemon's help. Doraemon gave a big light to both of them but the both big lights have different speed of magnifying. Let's assume the big light given to Gian can increase height of a person by v1 m/s and that of Suneo's big light is v2 m/s.
At the end of each second Doraemon check if their heights are equal or not.
Given initial height of Gian and Suneo, your task is to check whether the height of Gian and Suneo will become equal at some point or not, assuming they both started at the same time.First line takes the input of integer h1(height of gian), h2(height of suneo), v1(speed of Gian's big light) and v2(speed of Suneo's big light) as parameter.
<b>Constraints:-</b>
1 <b>≤</b> h2 < h1<b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v1 <b>≤</b> 10<sup>4</sup>
1 <b>≤</b> v2 <b>≤</b> 10<sup>4</sup>complete the function EqualOrNot and return a boolean True if their height will become equal at some point (as seen by Doraemon) else print False
Sample input:-
4 2 2 4
Sample output:-
Yes
Explanation:-
height of Gian goes as- 4 6 8 10. .
height of Suneo goes as:- 2 6 10..
at the end of 1 second their height will become equal.
Sample Input:-
5 4 1 6
Sample Output:
No, I have written this Solution Code: static boolean EqualOrNot(int h1, int h2, int v1,int v2){
if (v2>v1&&(h1-h2)%(v2-v1)==0){
return true;
}
return false;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, 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();
int i=str.length()-1;
if(i==0){
int number=Integer.parseInt(str);
System.out.println(number);
}else{
while(str.charAt(i)=='0'){
i--;
}
for(int j=i;j>=0;j--){
System.out.print(str.charAt(j));
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: n=int(input())
def reverse(n):
return int(str(n)[::-1])
print(reverse(n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a number, you task is to reverse its digits. If the reversed number contains 0s in the beginning, you must remove them as well.First line consists of a single integer denoting n.
Constraint:-
1<=size of n<=100Output is a single line containing reversed number n.Sample Input
123445
Sample Output
544321
Sample Input
16724368
Sample Output
86342761, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main()
{
string s;
cin>>s;
reverse(s.begin(),s.end());
int I;
for( I=0;I<s.length();I++){
if(s[I]!='0'){break;}
}
if(I==s.length()){cout<<0;return 0;}
for(int j=I;j<s.length();j++){
cout<<s[j];}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a one-dimensional sorted array A containing N integers.
You are also given an integer target, find if there exists a pair of elements in the array whose difference is equal to the target.
Approach this problem in O(n).The first line contains a single integer N.
The second line contains N space- separated integer A[i].
The third line contains an integer target.
Constraints
1<=N<=10^5
0<=A[i]<=10^9
0<=target<=10^9Print Yes if pair with given difference exists in our array otherwise print No.Sample Input 1:
5
1 2 7 9 11
5
Sample Output 1:
Yes
Sample Input 2:
5
1 1 8 8 25
0
Sample Output 2:
Yes, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
try{
int n = sc.nextInt();
int arr[] = new int[n];
HashMap<Integer, Integer> map =new HashMap<>();
for(int i=0; i<n; i++){
arr[i]=sc.nextInt();
map.put(arr[i], 1);
}
int target = sc.nextInt();
boolean found = false;
for(int i =0; i<n; i++){
if(map.containsKey(arr[i]+target)){
found=true;
break;
}
}
if(found)
System.out.println("Yes");
else
System.out.println("No");
}
catch(Exception e){
System.out.println("No");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a one-dimensional sorted array A containing N integers.
You are also given an integer target, find if there exists a pair of elements in the array whose difference is equal to the target.
Approach this problem in O(n).The first line contains a single integer N.
The second line contains N space- separated integer A[i].
The third line contains an integer target.
Constraints
1<=N<=10^5
0<=A[i]<=10^9
0<=target<=10^9Print Yes if pair with given difference exists in our array otherwise print No.Sample Input 1:
5
1 2 7 9 11
5
Sample Output 1:
Yes
Sample Input 2:
5
1 1 8 8 25
0
Sample Output 2:
Yes, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
bool check(vector<int> A,int target,int n) {
int i=0,j=1,x=0;
while(i<=n && j<=n)
{
x=A[j]-A[i];
if(x==target && i!=j)return 1;
else if(x<target)j++;
else i++;
}
return 0;
}
int main() {
int n,target;
cin>>n;
vector<int>arr(n);
for(int i=0;i<n;i++)cin>>arr[i];
cin>>target;
cout<<(check(arr,target,n)==1 ? "Yes" : "No");
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: static int isPresent(long arr[], int n, long k)
{
int left = 0;
int right = n-1;
int res = -1;
while(left<=right){
int mid = (left+right)/2;
if(arr[mid] == k){
res = 1;
break;
}else if(arr[mid] < k){
left = mid + 1;
}else{
right = mid - 1;
}
}
return res;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<long long,int> m;
long k;
cin>>k;
long long a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;
}
if(m.find(k)!=m.end()){
cout<<1<<endl;
}
else{
cout<<-1<<endl;
}
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: def binary_search(arr, low, high, x):
if high >= low:
mid = (high + low) // 2
if arr[mid] == x:
return 1
elif arr[mid] > x:
return binary_search(arr, low, mid - 1, x)
else:
return binary_search(arr, mid + 1, high, x)
else:
return -1
def position(n,arr,x):
return binary_search(arr,0,n-1,x)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a sorted array arr[] of N integers and a number K is given. The task is to check if the element K is present in the array or not.
<b>Note:</b> Use binary search to solve the problemThe first line of input contains a number of test cases T. For each test case, the first line of input contains a number of elements in the array, and the number K is separated by space. The next line contains N elements.
<b>Constraints:</b>
1 ≤ T ≤ 10
1 ≤ N ≤ 10<sup>5</sup>
1 ≤ K ≤ 10<sup>9</sup>
1 ≤ arr[i] ≤ 10<sup>9</sup>
<b>Sum of N over all test cases doesn't exceed 10<sup>6</sup></b>If the element is present in the array print "1" else print "-1".Sample Input:
2
5 6
1 2 3 4 6
5 2
1 3 4 5 6
Sample Output:
1
-1, I have written this Solution Code: // arr is they array to search from
// x is target
function binSearch(arr, x) {
// write code here
// do not console.log
// return the 1 or -1
let l = 0;
let r = arr.length - 1;
let mid;
while (r >= l) {
mid = l + Math.floor((r - l) / 2);
// If the element is present at the middle
// itself
if (arr[mid] == x)
return 1;
// If element is smaller than mid, then
// it can only be present in left subarray
if (arr[mid] > x)
r = mid - 1;
// Else the element can only be present
// in right subarray
else
l = mid + 1;
}
// We reach here when element is not
// present in array
return -1;
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix of characters. The matrix has N rows and M columns. Given a string s, you have to tell if it is possible to generate that string from given matrix.
Rules for generating string from matrix are:
You have to pick first character of string from row 1, second character from row 2 and so on. The (N+1)th character of string is to be picked from row 1, that is, you can traverse the rows in a cyclic manner (row 1 comes after row N).
If an occurrence of a character is picked from a row, you cannot pick the same occurrence again from that row.
You have to print Yes if given string can be generated from matrix using the given rules, else print No.First line consists of two integers N and M, denoting the matrix dimensions.
Next N lines consist of M characters each.
Last line consists of a string s.
Constraints:
1 ≤ N, M ≤ 200
1 ≤ S.length() ≤ 4*10<sup>4</sup>
S contains only lowercase English letters .
Print "Yes" if string can be generated else print "No". Answer for each test case should come in a new line.
Sample Input 1:
3 3
aba
xyz
bdr
axbaydb
Sample Output 1:
Yes
Explanation
We pick "a" from row 1. Now, we can only pick one more "a" from row 1 as one "a" is already used.
Similarly, "x" from row 2, "b" from row 3.
Now, we again go back to row 1.
We pick "a" from row 1, "y" from row 2 and so on., I have written this Solution Code: N,M = map(int,input().split())
words=[]
for i in range(N):
words.append(list(input().strip()))
test_word=list(input().strip())
ans="Yes"
for i in range(len(test_word)):
if test_word[i] not in words[i%N]:
ans="No"
break
else:
words[i%N].remove(test_word[i])
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a matrix of characters. The matrix has N rows and M columns. Given a string s, you have to tell if it is possible to generate that string from given matrix.
Rules for generating string from matrix are:
You have to pick first character of string from row 1, second character from row 2 and so on. The (N+1)th character of string is to be picked from row 1, that is, you can traverse the rows in a cyclic manner (row 1 comes after row N).
If an occurrence of a character is picked from a row, you cannot pick the same occurrence again from that row.
You have to print Yes if given string can be generated from matrix using the given rules, else print No.First line consists of two integers N and M, denoting the matrix dimensions.
Next N lines consist of M characters each.
Last line consists of a string s.
Constraints:
1 ≤ N, M ≤ 200
1 ≤ S.length() ≤ 4*10<sup>4</sup>
S contains only lowercase English letters .
Print "Yes" if string can be generated else print "No". Answer for each test case should come in a new line.
Sample Input 1:
3 3
aba
xyz
bdr
axbaydb
Sample Output 1:
Yes
Explanation
We pick "a" from row 1. Now, we can only pick one more "a" from row 1 as one "a" is already used.
Similarly, "x" from row 2, "b" from row 3.
Now, we again go back to row 1.
We pick "a" from row 1, "y" from row 2 and so on., I have written this Solution Code: import java.util.*;
import java.util.stream.*;
class Main {
public static void main(String args[] ) throws Exception {
Scanner sc = new Scanner(System.in);
int T = sc.nextInt();
for(int i = 0; i < T; i++)
{
int n = sc.nextInt();
int m = sc.nextInt();
String[][] arr = new String[n][1];
for (int j = 0; j < n; j++)
{
arr[j][0] = sc.next();
}
//System.out.println(Arrays.toString(arr[1])); o/p: [x, y, z]
//System.out.println(arr[1]); o/p: xyz
//System.out.println(arr[1].getClass().getSimpleName()); o/p:char[]
String s1 = sc.next();
int l = 0;
boolean ans = true;
while (s1.length() != l)
{
//String ls = Arrays.toString(arr[l%n]);
if (arr[l%n][0].contains(String.valueOf(s1.charAt(l))))
{
ans = true;
arr[l%n][0] = arr[l%n][0].replaceFirst(String.valueOf(s1.charAt(l)),"");
//arr[l%n][0] = ls;
}
else
{
ans = false;
break;
}
l+=1;
}
if (ans == true)
{
System.out.println("Yes");
}
else
{
System.out.println("No");
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
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