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For this Question: Given an integer N, your task is to check whether the given number is prime or notThe input contains a single integer N.
Constraints:-
1 <= N <= 100000000000Print "YES" If the given number is prime else print "NO".Sample Input:-
2
Sample Output:-
YES
Sample Input:-
4
Sample Output:-
NO, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
long n;
cin>>n;
long x = sqrt(n);
for(int i=2;i<=x;i++){
if(n%i==0){cout<<"NO";return 0;}
}
cout<<"YES";
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Olivia likes triangles. Hence, John decided to give her three integers and ask whether a triangle with edge length as given three integers is possible.
Help Olivia in answering it.The input consists of a single line containing three space-separated integers A, B, and C.
<b>Constraints </b>
1 <= A, B, C <= 100Output "Yes" if the triangle is possible otherwise, "No" (without quotes).Sample Input 1:
5 3 4
Sample Output 1:
Yes
Sample Explanation 1:
The possible triangle is a right-angled triangle with a hypotenuse of 5., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a, b, c;
cin >> a >> b >> c;
if(a<(b+c) && b<(c+a) && c<(a+b)){
cout << "Yes\n";
}else{
cout << "No\n";
}
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Alexa will turn on the air conditioner if, and only if, the temperature of the room is 30 degrees Celsius or above. The current temperature of the room is X degrees Celsius. Will she turn on the air conditioner?The input consists of a single integer X.
<b>Constraints</b>
−40≤X≤40
X is an integer.Print Yes if you will turn on the air conditioner; print No otherwise.<b>Sample Input 1</b>
25
<b>Sample Output 1</b>
No
<b>Sample Input 2</b>
30
<b>Sample Output 2</b>
Yes, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int x;
cin >> x;
if (x >= 30) cout << "Yes\n"; else cout << "No\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 <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: 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: Sara want to buy a toy and for this purpose she goes to a toy shop, there is a special offer going in the toy shop that is if you buy one toy you can get extra k toys for free now Sara wonders what is the minimum amount of money she need to spend so that she can get all the toys.
You are given some queries containing values of k you need to print the minimum amount of cost Sara need to spend to get all the toys for each value of k.First line contain number of toys N and the number of queries Q
Second line contains the cost of the toys
Next Q line contains a single integer that is the value of k
Constraint:-
1<=N, Q<=100000
1<=Arr[i], k<=1000000
Output the minimum cost for each querySample Input :
6 2
100 20 50 10 2 5
3
4
Sample Output :
7
7
Explanation:
She can buy toy at index 5 and get toys at index 1, 2, 3 for free than she buy toy at index 6 and get toy at index 4 for free so the total cost becomes 7 , I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args)throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] str = br.readLine().split(" ");
int N = Integer.parseInt(str[0]);
int Case = Integer.parseInt(str[1]);
long[] arr = new long[N];
String[] Str = br.readLine().split(" ");
for(int i = 0;i < N;i++){
arr[i] = Integer.parseInt(Str[i]);
}
Arrays.sort(arr);
for(int i = 1;i < N;i++){
arr[i] = arr[i-1] + arr[i];
}
double Q = 0;
while(Case-->0){
Q = Integer.parseInt(br.readLine());
int num = (int)Math.ceil(N/(Q+1));
System.out.println(arr[num-1]);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara want to buy a toy and for this purpose she goes to a toy shop, there is a special offer going in the toy shop that is if you buy one toy you can get extra k toys for free now Sara wonders what is the minimum amount of money she need to spend so that she can get all the toys.
You are given some queries containing values of k you need to print the minimum amount of cost Sara need to spend to get all the toys for each value of k.First line contain number of toys N and the number of queries Q
Second line contains the cost of the toys
Next Q line contains a single integer that is the value of k
Constraint:-
1<=N, Q<=100000
1<=Arr[i], k<=1000000
Output the minimum cost for each querySample Input :
6 2
100 20 50 10 2 5
3
4
Sample Output :
7
7
Explanation:
She can buy toy at index 5 and get toys at index 1, 2, 3 for free than she buy toy at index 6 and get toy at index 4 for free so the total cost becomes 7 , I have written this Solution Code: toy,qu = map(int,input().split())
cost = list(map(int,input().split()))
cost=sorted(cost)
for i in range(qu):
k=int(input())
l=len(cost)
t,s=0,0
while(l>0):
l=l-(k+1)
s=s+cost[t]
t=t+1
print(s,end="\n"), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara want to buy a toy and for this purpose she goes to a toy shop, there is a special offer going in the toy shop that is if you buy one toy you can get extra k toys for free now Sara wonders what is the minimum amount of money she need to spend so that she can get all the toys.
You are given some queries containing values of k you need to print the minimum amount of cost Sara need to spend to get all the toys for each value of k.First line contain number of toys N and the number of queries Q
Second line contains the cost of the toys
Next Q line contains a single integer that is the value of k
Constraint:-
1<=N, Q<=100000
1<=Arr[i], k<=1000000
Output the minimum cost for each querySample Input :
6 2
100 20 50 10 2 5
3
4
Sample Output :
7
7
Explanation:
She can buy toy at index 5 and get toys at index 1, 2, 3 for free than she buy toy at index 6 and get toy at index 4 for free so the total cost becomes 7 , I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define max1 10000001
int main(){
int n,q;
cin>>n>>q;
int a[n];
long b[n];
long sum=0;
for(int i=0;i<n;i++){
cin>>a[i];
}
sort(a,a+n);
for(int i=0;i<n;i++){
sum+=a[i];
b[i]=sum;
}
int k;
while(q--){
cin>>k;
int c = ceil(1.0*n/(k+1));
cout<<b[c-1]<<endl;
}
}
, 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 an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., 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());
if(n==1){
System.out.println(2);
}else{
int after = afterPrime(n);
int before = beforePrime(n);
if(before>after){
System.out.println(n+after);
}
else{System.out.println(n-before);}
}
}
public static boolean isPrime(int n)
{
int count=0;
for(int i=2;i*i<n;i++)
{
if(n%i==0)
count++;
}
if(count==0)
return true;
else
return false;
}
public static int beforePrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n-1;
c++;
}
}
}
public static int afterPrime(int n)
{
int c=0;
while(true)
{
if(isPrime(n))
return c;
else
{
n=n+1;
c++;
}
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., I have written this Solution Code: from math import sqrt
def NearPrime(N):
if N >1:
for i in range(2,int(sqrt(N))+1):
if N%i ==0:
return False
break
else: return True
else: return False
N=int(input())
i =0
while NearPrime(N-i)==False and NearPrime(N+i)==False:
i+=1
if NearPrime(N-i) and NearPrime(N+i):print(N-i)
elif NearPrime(N-i):print(N-i)
elif NearPrime(N+i): print(N+i), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N, find the closest prime number to N. If there are multiple print the smaller one.The input contains a single integer N.
Constraints:
1 <= N <= 1000000000Print the closest prime to N.Sample Input 1
12
Sample Output 1
11
Explanation: Closest prime to 12 is 11 and 13 smaller of which is 11.
Sample Input 2
17
Sample Output 2
17
Explanation: Closest prime to 17 is 17., 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>
/////////////
bool isPrime(int n){
if(n<=1)
return false;
for(int i=2;i*i<=n;++i)
if(n%i==0)
return false;
return true;
}
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
cin>>n;
if(n==1)
cout<<"2";
else{
int v1=n,v2=n;
while(isPrime(v1)==false)
--v1;
while(isPrime(v2)==false)
++v2;
if(v2-n==n-v1)
cout<<v1;
else{
if(v2-n<n-v1)
cout<<v2;
else
cout<<v1;
}
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, 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 str1 = br.readLine();
String[] str2 = str1.split(" ");
int[] arr = new int[n];
for(int i = 0; i < n; ++i) {
arr[i] = Integer.parseInt(str2[i]);
}
PermuteTheArray(arr, n);
}
static void PermuteTheArray(int A[], int n)
{
int []arr = new int[n];
for(int i = 0; i < n; i++)
{
arr[A[i] - 1] = i;
}
int mini = n, maxi = 0;
for(int i = 0; i < n; i++)
{
mini = Math.min(mini, arr[i]);
maxi = Math.max(maxi, arr[i]);
if (maxi - mini == i)
System.out.print(1+" ");
else
System.out.print(0+" ");
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, I have written this Solution Code: def find_permutations(arr):
max_ind = -1
min_ind = 10000000;
n = len(arr)
index_of = {}
for i in range(n):
index_of[arr[i]] = i + 1
for i in range(1, n + 1):
max_ind = max(max_ind, index_of[i])
min_ind = min(min_ind, index_of[i])
if (max_ind - min_ind + 1 == i):
print(1,end = " ")
else:
#print( "i = ",i)
print(0,end = " ")
n = int(input())
arr = list(map(int, input().split()))
find_permutations(arr), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara loves problems on permutation but this time she stuck on a problem and asks for your help.
Given a permutation of N integers as Arr[], your task is to check for each K(1 <= K <= N) there exists a subarray of size K such that it is also a permutation of K integers.
Note:- A permutation of N integers is a sequence of size N where every element from 1- N are present.The first line of input contains a single integer N denoting the size of permutation, the next line of input contains N space separated integers depicting the permutaiton.
Constraints:-
1 <= N <= 100000
1 <= Arr[i] <= NPrint N space separated integers either 1 or 0. Print 1 if there exist a permutation K for the ith number else print 0.Sample Input:-
6
4 5 1 3 2 6
Sample Output:-
1 0 1 0 1 1
Explanation:-
for k=1 permutaion exist from [3, 3]
for k=2 no permutaion exists
for k=3 permutaion exist from [3, 5]
for k=4 no permutaion exists
for k=5 permutaion exist from [1, 5]
for k=6 permutaion exist from [1, 6]
Sample Input:-
6
6 5 4 3 2 1
Sample Output:-
1 1 1 1 1 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 1001
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
pair<int,int> ans[100005];
signed main(){
// freopen("ou.txt", "r", stdin);
// freopen("output.txt", "w", stdout);
int t;t=1;
while(t--){
int n;
cin>>n;
int a[n+10];
int b[n+10];
FOR(i,n){
cin>>a[i];
b[a[i]]=i;
}
ans[1].first=b[1];
ans[1].second=b[1];
for(int i=2;i<=n;i++){
ans[i].first=min(ans[i-1].first,b[i]);
ans[i].second=max(ans[i-1].second,b[i]);
}
for(int i=1;i<=n;i++){
if(ans[i].second-ans[i].first+1==i){cout<<1<<" ";}
else{
cout<<0<<" ";
}
}
END;
}
}
, 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: You are given N intervals, the i<sup>th</sup> of them being [A<sub>i</sub>, B<sub>i</sub>], where A<sub>i</sub> and B<sub>i</sub> are positive integers. Let the union of all these intervals be S. It is easy to see that S can be uniquely represented as an union of disjoint closed intervals. Your task is to find the sum of the lengths of the disjoint closed intervals that comprises S.
For example, if you are given the intervals: [1, 3], [2, 4], [5, 7] and [7, 8], then S can be uniquely represented as the union of disjoint intervals [1, 4] and [5, 8]. In this case, the answer will be 6, as (4 - 1) + (8 - 5) = 6.The first line of the input consists of a single integer N – the number of intervals.
Then N lines follow, the i<sup>th</sup> line containing two space-separated integers A<sub>i</sub> and B<sub>i</sub>.
<b> Constraints: </b>
1 ≤ N ≤ 10<sup>4</sup>
1 ≤ A<sub>i</sub> < B<sub>i</sub> ≤ 2×10<sup>5</sup>Print a single integer, the sum of the lengths of the disjoint intervals of S.Sample Input 1:
3
1 3
3 4
6 7
Sample Output 1:
4
Sample Explanation 1:
Here, S can be uniquely written as union of disjoint intervals [1, 4] and [6, 7]. Thus, answer is (4 - 1) + (7 - 6) = 4.
Sample Input 2:
4
9 12
8 10
5 6
13 15
Sample Output 2:
7
Sample Explanation 2:
Here, S can be uniquely written as union of disjoint intervals [5, 6], [8, 12] and [13, 15]. Thus, answer is (6 - 5) + (12 - 8) + (15 - 13) = 7.
, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << (a) << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (auto i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 998244353; // 1e9 + 7;
const ll N = 4e5 + 100;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
int arr[N];
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
read(n);
FOR (i, 1, n)
{
readb(a, b);
a *= 2, b *= 2;
arr[a]++;
arr[b + 1]--;
}
int ans = 0, sum = 0, cur = 0;
FOR (i, 0, N - 2)
{
sum += arr[i];
if (sum) cur++;
else ans += cur/2, cur = 0;
}
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String s = in.readLine().trim();
char[] express = s.toCharArray();
StringBuilder ans = new StringBuilder("");
Stack<Character> stack = new Stack<>();
for(int i=0; i<express.length; i++){
if(express[i]>=65 && express[i]<=90){
ans.append(express[i]);
}else if(express[i]==')'){
while(stack.peek()!='('){
ans.append(stack.peek());
stack.pop();
}
stack.pop();
}else if(stack.empty() || express[i]=='(' || precedence(express[i])>=precedence(stack.peek())){
stack.push(express[i]);
}else{
while(!stack.empty() && stack.peek()!='(' && precedence(stack.peek())<=express[i]){
ans.append(stack.peek());
stack.pop();
}
stack.push(express[i]);
}
}
while(!stack.empty()){
ans.append(stack.peek());
stack.pop();
}
System.out.println(ans);
}
static int precedence(char operator){
int precedenceValue;
if(operator=='+' || operator=='-')
precedenceValue = 1;
else if(operator=='*' || operator=='/')
precedenceValue = 2;
else if(operator == '^')
precedenceValue = 3;
else
precedenceValue = -1;
return precedenceValue;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: string=input()
stack=list()
preced={'+':1,'-':1,'*':2,'/':2,'^':3}
for i in string:
if i!=')':
if i=='(':
stack.append(i)
elif i in ['*','+','/','-','^']:
while(stack and stack[-1]!='(' and preced[i]<=preced[stack[-1]]):
print(stack.pop(),end="")
stack.append(i)
else:
print(i,end="")
else:
while(stack and stack[-1]!='('):
print(stack.pop(),end="")
stack.pop()
while(stack):
print(stack.pop(),end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
//Function to return precedence of operators
int prec(char c)
{
if(c == '^')
return 3;
else if(c == '*' || c == '/')
return 2;
else if(c == '+' || c == '-')
return 1;
else
return -1;
}
// The main function to convert infix expression
//to postfix expression
void infixToPostfix(string s)
{
std::stack<char> st;
st.push('N');
int l = s.length();
string ns;
for(int i = 0; i < l; i++)
{
// If the scanned character is an operand, add it to output string.
if((s[i] >= 'a' && s[i] <= 'z')||(s[i] >= 'A' && s[i] <= 'Z'))
ns+=s[i];
// If the scanned character is an ‘(‘, push it to the stack.
else if(s[i] == '(')
st.push('(');
// If the scanned character is an ‘)’, pop and to output string from the stack
// until an ‘(‘ is encountered.
else if(s[i] == ')')
{
while(st.top() != 'N' && st.top() != '(')
{
char c = st.top();
st.pop();
ns += c;
}
if(st.top() == '(')
{
char c = st.top();
st.pop();
}
}
//If an operator is scanned
else{
while(st.top() != 'N' && prec(s[i]) <= prec(st.top()))
{
char c = st.top();
st.pop();
ns += c;
}
st.push(s[i]);
}
}
//Pop all the remaining elements from the stack
while(st.top() != 'N')
{
char c = st.top();
st.pop();
ns += c;
}
cout << ns << endl;
}
//Driver program to test above functions
int main()
{
string exp;
cin>>exp;
infixToPostfix(exp);
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code:
#include <iostream>
using namespace std;
int Dishes(int N, int T){
return T-N;
}
int main(){
int n,k;
scanf("%d%d",&n,&k);
printf("%d",Dishes(n,k));
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sheldon and Leonard are gone for lunch but none of them have money so they decided to wash dishes. In total, they washed T dishes from which N dishes are washed by Leonard. Now Leonard wants to know the number of dishes Sheldon washed. Help him to find it.The first line of the input contains N and T
Constraints:-
1 <= N <= T <= 1000Return the number of dishes Sheldon washed.Sample Input:-
3 6
Sample Output:-
3
Sample Input:-
2 4
Sample Output:-
2, I have written this Solution Code: nan, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) {
try{
InputStreamReader r=new InputStreamReader(System.in);
BufferedReader in = new BufferedReader(r);
String a = in.readLine();
String[] nums = a.split(" ");
long[] l = new long[3];
for(int i=0; i<3; i++){
l[i] = Long.parseLong(nums[i]);
}
Arrays.sort(l);
System.out.print(l[1]);
}
catch(Exception e){
System.out.println(e);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code:
//#define ASC
//#define DBG_LOCAL
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace std;
using namespace __gnu_pbds;
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// #pragma GCC optimize("Ofast")
// #pragma GCC target("avx,avx2,fma")
#define int long long
// #define int __int128
#define all(X) (X).begin(), (X).end()
#define pb push_back
#define endl '\n'
#define fi first
#define se second
// const int mod = 1e9 + 7;
const int mod=998'244'353;
const long long INF = 2e18 + 10;
// const int INF=1e9+10;
#define readv(x, n) \
vector<int> x(n); \
for (auto &i : x) \
cin >> i;
template <typename T>
using v = vector<T>;
template <typename T>
using vv = vector<vector<T>>;
template <typename T>
using vvv = vector<vector<vector<T>>>;
typedef vector<int> vi;
typedef vector<double> vd;
typedef vector<vector<int>> vvi;
typedef vector<vector<vector<int>>> vvvi;
typedef vector<vector<vector<vector<int>>>> vvvvi;
typedef vector<vector<double>> vvd;
typedef pair<int, int> pii;
int multiply(int a, int b, int in_mod) { return (int)(1LL * a * b % in_mod); }
int mult_identity(int a) { return 1; }
const double PI = acosl(-1);
auto power(auto a, auto b, const int in_mod)
{
auto prod = mult_identity(a);
auto mult = a % 2;
while (b != 0)
{
if (b % 2)
{
prod = multiply(prod, mult, in_mod);
}
if(b/2)
mult = multiply(mult, mult, in_mod);
b /= 2;
}
return prod;
}
auto mod_inv(auto q, const int in_mod)
{
return power(q, in_mod - 2, in_mod);
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define stp cout << fixed << setprecision(20);
void solv()
{
int A ,B, C;
cin>>A>>B>>C;
vector<int> values;
values.push_back(A);
values.push_back(B);
values.push_back(C);
sort(all(values));
cout<<values[1]<<endl;
}
void solve()
{
int t = 1;
// cin>>t;
for(int i = 1;i<=t;i++)
{
// cout<<"Case #"<<i<<": ";
solv();
}
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(NULL);
cerr.tie(NULL);
#ifndef ONLINE_JUDGE
if (fopen("INPUT.txt", "r"))
{
freopen("INPUT.txt", "r", stdin);
freopen("OUTPUT.txt", "w", stdout);
}
#else
#ifdef ASC
namespace fs = std::filesystem;
std::string path = "./";
string filename;
for (const auto & entry : fs::directory_iterator(path)){
if( entry.path().extension().string() == ".in"){
filename = entry.path().filename().stem().string();
}
}
if(filename != ""){
string input_file = filename +".in";
string output_file = filename +".out";
if (fopen(input_file.c_str(), "r"))
{
freopen(input_file.c_str(), "r", stdin);
freopen(output_file.c_str(), "w", stdout);
}
}
#endif
#endif
// auto clk = clock();
// -------------------------------------Code starts here---------------------------------------------------------------------
signed t = 1;
// cin >> t;
for (signed test = 1; test <= t; test++)
{
// cout<<"Case #"<<test<<": ";
// cout<<endl;
solve();
}
// -------------------------------------Code ends here------------------------------------------------------------------
// clk = clock() - clk;
#ifndef ONLINE_JUDGE
// cerr << fixed << setprecision(6) << "\nTime: " << ((float)clk) / CLOCKS_PER_SEC << "\n";
#endif
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given three distinct positive integers, A, B and C, in the input. Your task is to find their median.
The median of a set of integers is the number at the middle position if they are arranged in ascending order.The input consists of a single line containing three integers A, B and C.
<b> Constraints: </b>
1 ≤ A, B, C ≤ 10<sup>9</sup>
A, B, C are pairwise distinct.
Print a single integer, the median of the given numbers.Sample Input 1:
5 1 3
Sample Output 1:
3
Sample Input 2:
1 2 3
Sample Output 2:
2, I have written this Solution Code: lst = list(map(int, input().split()))
lst.sort()
print(lst[1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: import java.io.*;
import java.util.*;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
class Main {
public static void main (String[] args)
throws IOException
{
BufferedReader read = new BufferedReader(new InputStreamReader(System.in));
String str = read.readLine();
System.out.println(maxProduct(str));
}
public static long maxProduct(String str) {
StringBuilder sb = new StringBuilder(str);
int x = sb.length();
int[] dpl = new int[x];
int[] dpr = new int[x];
modifiedOddManacher(sb.toString(), dpl);
modifiedOddManacher(sb.reverse().toString(), dpr);
long max=1;
for(int i=0;i<x-1;i++)
max=Math.max(max, (1+(dpl[i]-1)*2L)*(1+(dpr[x-(i+1)-1]-1)*2L));
return max;
}
private static void modifiedOddManacher(String str, int[] dp){
int x = str.length();
int[] center = new int[x];
for(int l=0,r=-1,i=0;i<x;i++){
int radius = (i > r) ? 1 : Math.min(center[l+(r-i)], r-i+1);
while(i-radius>=0 && i+radius<x && str.charAt(i-radius)==str.charAt(i+radius)) {
dp[i+radius] = radius+1;
radius++;
}
center[i] = radius--;
if(i+radius>r){
l = i-radius;
r = i+radius;
}
}
for(int i=0, max=1;i<x;i++){
max = Math.max(max, dp[i]);
dp[i] = max;
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: s=input()
n = len(s)
hlen = [0]*n
center = right = 0
for i in range(n):
if i < right:
hlen[i] = min(right - i, hlen[2*center - i])
while 0 <= i-1-hlen[i] and i+1+hlen[i] < len(s) and s[i-1-hlen[i]] == s[i+1+hlen[i]]:
hlen[i] += 1
if right < i+hlen[i]:
center, right = i, i+hlen[i]
left = [0]*n
right = [0]*n
for i in range(n):
left[i+hlen[i]] = max(left[i+hlen[i]], 2*hlen[i]+1)
right[i-hlen[i]] = max(right[i-hlen[i]], 2*hlen[i]+1)
for i in range(1, n):
left[~i] = max(left[~i], left[~i+1]-2)
right[i] = max(right[i], right[i-1]-2)
for i in range(1, n):
left[i] = max(left[i-1], left[i])
right[~i] = max(right[~i], right[~i+1])
print(max(left[i-1]*right[i] for i in range(1, n))), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string of length N. You have to select two non- overlapping (no common index) non- empty substrings of odd lengths from that string such that both those substrings are palindrome. You want the product of lengths of those substring to be maximum.Input contains of a single String of length N.
Constraints
2 <= N <= 100000
String contains lowercase english letters.Print a single integer which is the maximum possible product of lengths of selected substrings.Sample input 1
aabaaba
Sample output 1
9
Explanation : we can select substring [2-4] = aba and [5-7] = aba the product of their lengths is 9.
Sample Input 2
aabababaaa
Sample Output 2
15
, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountl
#define m_p make_pair
#define inf 200000000000000
#define MAXN 1000001
#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);
}
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
#define S second
#define F first
#define int long long
/////////////
int v1[100001]={};
int v2[100001]={};
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n;
string s;
cin>>s;
n=s.length();
vector<int> d1(n);
for (int i = 0, l = 0, r = -1; i < n; i++) {
int k = (i > r) ? 1 : min(d1[l + r - i], r - i + 1);
while (0 <= i - k && i + k < n && s[i - k] == s[i + k]) {
k++;
}
d1[i] = k--;
if (i + k > r) {
l = i - k;
r = i + k;
}
}
int c=0;
for(int i=0;i<n;++i)
{
int x=2*d1[i]-1;
int j=i+d1[i]-1;
while(v1[j]<x&&j>=i)
{
v1[j]=x;
x-=2;
++c;
--j;
}
}
for(int i=1;i<n;++i)
{
v1[i]=max(v1[i],v1[i-1]);
}
for(int i=n-1;i>=0;--i)
{
int x=2*d1[i]-1;
int j=i-d1[i]+1;
while(v2[j]<x&&j<=i)
{
v2[j]=x;
x-=2;
++j;
++c;
}
}
for(int i=n-2;i>=0;--i)
{
v2[i]=max(v2[i],v2[i+1]);
}
int ans=0;
for(int i=1;i<n;++i)
{
ans=max(ans,v1[i-1]*v2[i]);
}
cout<<ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, 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[][]arr=new int[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
arr[i][j]=sc.nextInt();
}
}
int determinent=(arr[0][0]*arr[1][1])-(arr[1][0]*arr[0][1]);
System.out.print(determinent);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
lis=[]
a11=arr1[0]
a12=arr1[1]
a21=arr2[0]
a22=arr2[1]
det=(a11*a22)-(a12*a21)
print(det), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<a*d-b*c;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
String s = in.readLine().trim();
char[] express = s.toCharArray();
StringBuilder ans = new StringBuilder("");
Stack<Character> stack = new Stack<>();
for(int i=0; i<express.length; i++){
if(express[i]>=65 && express[i]<=90){
ans.append(express[i]);
}else if(express[i]==')'){
while(stack.peek()!='('){
ans.append(stack.peek());
stack.pop();
}
stack.pop();
}else if(stack.empty() || express[i]=='(' || precedence(express[i])>=precedence(stack.peek())){
stack.push(express[i]);
}else{
while(!stack.empty() && stack.peek()!='(' && precedence(stack.peek())<=express[i]){
ans.append(stack.peek());
stack.pop();
}
stack.push(express[i]);
}
}
while(!stack.empty()){
ans.append(stack.peek());
stack.pop();
}
System.out.println(ans);
}
static int precedence(char operator){
int precedenceValue;
if(operator=='+' || operator=='-')
precedenceValue = 1;
else if(operator=='*' || operator=='/')
precedenceValue = 2;
else if(operator == '^')
precedenceValue = 3;
else
precedenceValue = -1;
return precedenceValue;
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: string=input()
stack=list()
preced={'+':1,'-':1,'*':2,'/':2,'^':3}
for i in string:
if i!=')':
if i=='(':
stack.append(i)
elif i in ['*','+','/','-','^']:
while(stack and stack[-1]!='(' and preced[i]<=preced[stack[-1]]):
print(stack.pop(),end="")
stack.append(i)
else:
print(i,end="")
else:
while(stack and stack[-1]!='('):
print(stack.pop(),end="")
stack.pop()
while(stack):
print(stack.pop(),end=""), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an infix expression, your task is to convert it into postfix.
<b>Infix expression</b>: The expression of the form a operator b. When an operator is in- between every pair of operands.
<b>Postfix expression</b>: The expression of the form a b operator. When an operator is followed for every pair of operands.
Valid operators are +, - , *, /, ^. Each operand is an uppercase english alphabet. Infix expression may contains parenthesis as '(' and ')'.
See example for better understanding.The input contains a single string of infix expression.
<b>Constraints:-</b>
1 <= |String| <= 40Output the Postfix expression.Sample Input:-
((A-(B/C))*((A/K)-L))
Sample Output:-
ABC/-AK/L-*
Sample Input:-
A+B
Sample Output:-
AB+, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
//Function to return precedence of operators
int prec(char c)
{
if(c == '^')
return 3;
else if(c == '*' || c == '/')
return 2;
else if(c == '+' || c == '-')
return 1;
else
return -1;
}
// The main function to convert infix expression
//to postfix expression
void infixToPostfix(string s)
{
std::stack<char> st;
st.push('N');
int l = s.length();
string ns;
for(int i = 0; i < l; i++)
{
// If the scanned character is an operand, add it to output string.
if((s[i] >= 'a' && s[i] <= 'z')||(s[i] >= 'A' && s[i] <= 'Z'))
ns+=s[i];
// If the scanned character is an ‘(‘, push it to the stack.
else if(s[i] == '(')
st.push('(');
// If the scanned character is an ‘)’, pop and to output string from the stack
// until an ‘(‘ is encountered.
else if(s[i] == ')')
{
while(st.top() != 'N' && st.top() != '(')
{
char c = st.top();
st.pop();
ns += c;
}
if(st.top() == '(')
{
char c = st.top();
st.pop();
}
}
//If an operator is scanned
else{
while(st.top() != 'N' && prec(s[i]) <= prec(st.top()))
{
char c = st.top();
st.pop();
ns += c;
}
st.push(s[i]);
}
}
//Pop all the remaining elements from the stack
while(st.top() != 'N')
{
char c = st.top();
st.pop();
ns += c;
}
cout << ns << endl;
}
//Driver program to test above functions
int main()
{
string exp;
cin>>exp;
infixToPostfix(exp);
return 0;
} , In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Akash is hosting a farewell party and wants to serve the most delicious pizzas to his K friends. There are N restaurants in the city, numbered from 1 to N, and each restaurant has a rating between 1 and 100. Akash wants to select K different restaurants such that the minimum rating of any selected restaurant is maximized. Help Akash find the maximum possible minimum rating of the K-selected restaurants for his farewell party.The first line of the input contains 2 space-separated integers N and K.
The second line of the input contains N space-separated integers denoting the rating of restaurants.
<b>Constraints</b>
1 ≤ K ≤ N ≤ 10<sup>5</sup>Print the maximum possible minimum rating of the K selected restaurants for Akash's farewell party.<b>Sample Input</b>
5 3
10 20 30 40 50
<b>Sample output</b>
30
<b>Explanation</b>
Akash can select restaurants with ratings 30, 40, and 50 for his friends. The minimum rating among these restaurants is 30, which is the maximum possible minimum rating of the selected restaurants., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
vector<int> a(n);
for (int i = 0; i < n; i++) {
cin >> a[i];
}
sort(a.begin(), a.end());
cout << a[n - k] << "\n";
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader bd=new BufferedReader(new InputStreamReader(System.in));
String st1=bd.readLine();
int n=Integer.parseInt(st1);
int arr[]=new int[n];
String[] st2=bd.readLine().split(" ");
for(int i=0; i<n; i++){
arr[i]=Integer.parseInt(st2[i]);
}
long sum_till_here_forward = 0;
long max_sum_forward = 0;
long sum_till_here_backward = 0;
long max_sum_backward = 0;
long sum_forward[] = new long[n+1];
long sum_backward[] = new long[n+1];
long maximum=0;
if(n==2){
maximum=Math.max(arr[0],arr[1]);
}
else{
for(int i=0; i<n; i++){
sum_till_here_forward += arr[i];
if(sum_till_here_forward > max_sum_forward){
max_sum_forward = sum_till_here_forward;
}
if(sum_till_here_forward < 0){
sum_till_here_forward = 0;
}
sum_forward[i+1] = max_sum_forward;
}
for(int i=n-1; i>=0; i--){
sum_till_here_backward += arr[i];
if(sum_till_here_backward > max_sum_backward){
max_sum_backward = sum_till_here_backward;
}
if(sum_till_here_backward < 0){
sum_till_here_backward=0;
}
sum_backward[i+1] = max_sum_backward;
}
for(int i=1; i<n; i++){
maximum=Math.max(maximum,(sum_forward[i-1]+sum_backward[i+1]));
}
}
System.out.print(maximum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: n=int(input())
l=list(map(int,input().split()))
left=[]
for i in range(n):
left.append(0)
curr_max=l[0]
max_so_far=l[0]
for i in range(1,n):
left[i]=max_so_far
curr_max = max(l[i], curr_max+l[i])
max_so_far = max(max_so_far, curr_max)
right=[]
for i in range(n):
right.append(0)
curr_max=l[n-1]
max_so_far=l[n-1]
for i in range(n-2,-1,-1):
right[i]=max_so_far
curr_max = max(l[i], curr_max+l[i])
max_so_far = max(max_so_far, curr_max)
ans=0
for i in range(n):
ans=max(ans,left[i]+right[i])
print(ans), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: After sitting idle in home for around 2 months, Jiraya has finally come up with a super interesting problem on arrays (or maybe!).
You are given an array A of N integers. You need to choose an index i from 1 to N (inclusive) and divide this array into 2 parts, then find the maximum sum of contiguous subarray in both the parts of arrays and add the obtained values from the two arrays.
What is the maximum sum that you can obtain?
Note: The element at index i is not a part of any of the generated arrays, choosing an empty subarray is allowed, one of the generated array is empty if i=1 or i=N (its maximum subarray sum will be 0).
See sample for better understanding.The first line of input contains a single integer N.
The second line of input contains N integers A[1], A[2],. , A[N]
Constraints
2 <= N <= 200000
-1000000 <= A[i] <= 1000000Output a single integer, the answer to the above problem.
(The answer may not fit into integer data type)Sample Input
6
-5 -1 4 -3 5 -4
Sample Output
9
Explanation: We choose i = 4. The two arrays are [-5, -1, 4] and [5, -4]. The maximum sum of contiguous subarrays are 4 and 5 respectively. Therefore, the answer is 4 + 5 = 9.
Sample Input
5
-1 -1 -1 -1 -1
Sample Output
0
Explanation: We choose i = 3. The two arrays are [-1, -1] and [-1, -1]. The maximum sum of contiguous subarrays are 0 and 0 respectively (empty subarrays). Therefore, the answer is 0 + 0 = 0., I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define sd(x) scanf("%d", &x)
#define sz(v) (int) v.size()
#define pr(v) For(i, 0, sz(v)) {cout<<v[i]<<" ";} cout<<endl;
#define slld(x) scanf("%lld", &x)
#define all(x) x.begin(), x.end()
#define For(i, st, en) for(ll i=st; i<en; i++)
#define tr(x) for(auto it=x.begin(); it!=x.end(); it++)
#define fast std::ios::sync_with_stdio(false);cin.tie(NULL);
#define pb push_back
#define ll long long
#define int long long
#define ld long double
#define mp make_pair
#define F first
#define S second
typedef pair<int, int> pii;
typedef vector<int> vi;
#define pi 3.141592653589793238
#define MOD 1000000007
#define INF 1000000000000000007LL
const int N = 200005;
// it's swapnil07 ;)
#ifdef SWAPNIL07
#define trace(...) __f(#__VA_ARGS__, __VA_ARGS__)
template <typename Arg1>
void __f(const char* name, Arg1&& arg1){
cout << name << " : " << arg1 << endl;
}
template <typename Arg1, typename... Args>
void __f(const char* names, Arg1&& arg1, Args&&... args){
const char* comma = strchr(names + 1, ',');cout.write(names, comma - names) << " : " << arg1<<" | ";__f(comma+1, args...);
}
int begtime = clock();
#define end_routine() cout << "\n\nTime elapsed: " << (clock() - begtime)*1000/CLOCKS_PER_SEC << " ms\n\n";
#else
#define endl '\n'
#define trace(...)
#define end_routine()
#endif
vector<int> a;
int n;
vector<int> kadane(){
int cur = 0;
int mx = 0;
vector<int> vect(n);
for(int i=0; i<n; i++){
cur += a[i];
mx = max(mx, cur);
if(cur < 0)
cur = 0;
vect[i]=mx;
}
return vect;
}
signed main()
{
fast
#ifdef SWAPNIL07
freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);
#endif
cin>>n;
for(int i=0; i<n; i++){
int x; cin>>x;
a.pb(x);
}
vector<int> v1 = kadane();
reverse(all(a));
vector<int> v2 = kadane();
reverse(all(v2));
int ans = 0;
For(i, 0, n){
if(i>0 && i<n-1)
ans = max(ans, v1[i-1]+v2[i+1]);
else if(i==0)
ans = max(ans, v2[i+1]);
else
ans = max(ans, v1[i-1]);
}
cout<<ans;
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 1, I have written this Solution Code: n=int(input())
x=n/3
if n%3==2:
x+=1
print(int(x)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 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 ans = n/3;
if(n%3==2){ans++;}
System.out.print(ans);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: In this season of love, everyone wants to surprise each other.
You are also super excited and you wish to buy roses of 3 different colors. You always buy roses in order, white, yellow, red.
So if you buy 7 roses, they will be "white, yellow, red, white, yellow, red, white".
You need to find the number of yellow roses that you will buy?The only line of input contains a single integer, N, the number of roses that you will buy.
Constraints
1 <= N <= 1000Output a single integer, the number of yellow roses.Sample Input 1
2
Sample Output 1
1
Sample Input 2
8
Sample Ouput 2
3
Explanation;-
testcase1;- 2 flower will be white,yellow
so number of yellow flower is 1, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int n;
cin>>n;
int x=n/3;
if(n%3==2){
x++;}
cout<<x;}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
BufferedReader in =new BufferedReader(new InputStreamReader(System.in));
StringBuilder s = new StringBuilder();
String text=null;
while ((text = in.readLine ()) != null)
{
s.append(text);
}
int len=s.length();
for(int i=0;i<len-1;i++){
if(s.charAt(i)==s.charAt(i+1)){
int flag=0;
s.delete(i,i+2);
int left=i-1;
len=len-2;
i=i-2;
if(i<0){
i=-1;
}
}
}
System.out.println(s);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: s=input()
l=["aa","bb","cc","dd","ee","ff","gg","hh","ii","jj","kk","ll","mm","nn","oo","pp","qq","rr","ss","tt","uu","vv","ww","xx","yy","zz"]
while True:
do=False
for i in range(len(l)):
if l[i] in s:
do=True
while l[i] in s:
s=s.replace(l[i],"")
if do==False:
break
print(s), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S you have to remove all pairs of identical letters, which are consecutive. If after the removal there appear new pairs, remove them as well. Repeat this untill no adjacent letter in the string is same.
Note that deleting of the consecutive identical letters can be done in any order, as any order leads to the same result.The input data consists of a single string S.
Constraints:
1 <= |S| <= 100000
S contains lowercase english letters only.Print the given string after it is processed. It is guaranteed that the result will contain at least one character.Sample Input
hhoowaaaareyyoouu
Sample Output
wre
Explanation:
First we remove "hh" then "oo" then "aa" then "yy" then "oo" then "uu" and we are left with "wre".
Now we cannot remove anything.
Sample Input:-
abcde
Sample Output:-
abcde
Sample Input:-
abcddcb
Sample Output:-
a, I have written this Solution Code: #pragma GCC optimize ("Ofast")
#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
template<class T>
using oset = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main(){
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
string s;
cin>>s;
int len=s.length();
char stk[410000];
int k = 0;
for (int i = 0; i < len; i++)
{
stk[k++] = s[i];
while (k > 1 && stk[k - 1] == stk[k - 2])
k -= 2;
}
for (int i = 0; i < k; i++)
cout << stk[i];
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, 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 str[]=br.readLine().split(" ");
int a[]=new int[n];
for(int i=0;i<n;i++){
a[i]=Integer.parseInt(str[i]);
}
long res=0;
for (int i=0;i<32;i++){
long cnt=0;
for (int j=0;j<n;j++)
if ((a[j] & (1 << i)) == 0)
cnt++;
res=(res+(cnt*(n-cnt)*2))%1000000007;
}
System.out.println(res%1000000007);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, I have written this Solution Code: def suBD(arr, n):
ans = 0 # Initialize result
for i in range(0, 64):
count = 0
for j in range(0, n):
if ( (arr[j] & (1 << i)) ):
count+= 1
ans += (count * (n - count)) * 2;
return (ans)%(10**9+7)
n=int(input())
arr = map(int,input().split())
arr=list(arr)
print(suBD(arr, n)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an array of N integers, your task is to calculate the sum of bit difference in all pairs which can be formed.The first line of input contains a single integer N, the second line of input contains N space separated integers depicting values of the array.
Constraints:-
1 <= N <= 100000
0 <= Arr[i] <= 1000000000Print the sum of bit difference of all possible pairs.
Note:- Since the answer can be quite large print your answer modulo 10<sup>9</sup> + 7Sample Input:-
2
1 3
Sample Output:-
2
Explanation:-
(1, 1) = 0
(1, 3) = 1
(3, 1) = 1
(3, 3) = 0
Sample Input:-
2
1 2
Sample Output:-
4, I have written this Solution Code:
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define max1 101
#define MOD 1000000007
#define read(type) readInt<type>()
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
#define int long long
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}signed main(){
int N;
cin>>N;
int a[55];
int A[N];
FOR(i,N){
cin>>A[i];}
for(int i=0;i<55;i++){
a[i]=0;
}
int ans=1,p=2;
for(int i=0;i<55;i++){
for(int j=0;j<N;j++){
if(ans&A[j]){a[i]++;}
}
ans*=p;
// out(ans);
}
ans=0;
for(int i=0;i<55;i++){
ans+=(a[i]*(N-a[i])*2);
ans%=MOD;
}
out(ans);
}
, 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: You are given an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String[] strArr = br.readLine().split(" ");
int N = Integer.parseInt(strArr[0]);
int Q = Integer.parseInt(strArr[1]);
int[] arr = new int[100001];
strArr = br.readLine().split(" ");
int max = 0;
for(int i=0; i<N; i++){
int curr = Integer.parseInt(strArr[i]);
arr[curr]++;
if(curr>max){
max = curr;
}
}
long[] prefixSum = new long[100001];
long sum = 0;
for(int i=0; i<=100000; i++){
sum = sum + arr[i];
prefixSum[i] = sum;
}
for(int i=0; i<Q; i++){
strArr = br.readLine().split(" ");
int L = Integer.parseInt(strArr[0]);
int R = Integer.parseInt(strArr[1]);
long count = prefixSum[R]-prefixSum[L-1];
System.out.println(count);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, I have written this Solution Code: n,q=map(int,input().split())
a=list(map(int,input().split()))
b=[0]*(100001)
for i in range(n):
b[a[i]]+=1
s=0
for i in range(100001):
s+=b[i]
b[i]=s
for i in range(q):
v=[int(k) for k in input().split()]
print(b[v[1]]-b[v[0]-1]), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an array Arr of N integers. You have to process Q queries on the array. Each query contains L and R, for each query you have to report the number of indices i in the array such that L <= Arr[i] <= R.First line of input contains two integers N and Q representing number of elements in the array and number of queries.
Second line of input contains N integers representing the array Arr.
Next Q lines of input contains L and R for that query.
Constraints
1 <= N,Q <= 100000
1 <= Arr[i] <= 100000
1 <= L <= R <= 100000For each query you have to print the number of index i in the array such that L <= Arr[i] <= R in a seperate line.Sample input
10 10
3 6 7 4 9 1 7 10 9 4
1 9
4 6
4 7
4 6
6 8
5 8
2 6
2 3
9 9
2 5
Sample output
9
3
5
3
3
3
4
1
2
3
Explanation :
For range 1 to 9 : all indexes except index 8 (with value 10) have values in the range 1 to 9
For range 4 to 6 : indexes 2(with value 6), 4(with value 4) and 10(with value 4) have value in the range 4 to 6, I have written this Solution Code: #include<bits/stdc++.h>
using namespace std;
#define ll long long
#define VV vector
#define pb push_back
#define bitc __builtin_popcountll
#define m_p make_pair
#define infi 1e18+1
#define eps 0.000000000001
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);
string char_to_str(char c){string tem(1,c);return tem;}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
template<class T>//usage rand<long long>()
T rand() {
return uniform_int_distribution<T>()(rng);
}
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
typedef tree<int, null_type, less<int>, rb_tree_tag, tree_order_statistics_node_update> oset;
// string to integer stoi()
// string to long long stoll()
// string.substr(position,length);
// integer to string to_string();
//////////////
auto clk=clock();
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define sz(x) ((long long)x.size())
#define int long long
#define f80 __float128
#define pii pair<int,int>
/////////////
signed main()
{
fastio;
#ifdef ANIKET_GOYAL
freopen("inputf.in","r",stdin);
freopen("outputf.in","w",stdout);
#endif
int n,q;
cin>>n>>q;
int c[100001]={};
int a[n+1];
for(int i=1;i<=n;++i)
{
cin>>a[i];
c[a[i]]++;
}
vector<int> ans;
for(int i=1;i<=100000;++i)
c[i]=c[i]+c[i-1];
for(int i=1;i<=q;++i)
{
int l,r;
cin>>l>>r;
ans.push_back(c[r]-c[l-1]);
}
for(auto r:ans)
cout<<r<<"\n";
#ifdef ANIKET_GOYAL
// cout<<endl<<endl<<endl<<endl<<"Time elapsed: "<<(double)(clock()-clk)/CLOCKS_PER_SEC<<endl;
#endif
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code:
int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: C, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: static int mythOrFact(int N){
int prime =1;
int cnt = N+1;
for(int i=2;i<N;i++){
if(N%i==0){cnt+=i;prime=0;}
}
int p = 3*N;
p/=2;
if((cnt<=p && prime==1) || (cnt>p && prime==0) ){return 1;}
return 0;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There is a saying in Sara's village that "A number is called prime if the sum of all the factors of N is less than or equal to (3*N)/2 ". Given the number N, your task is to check if it is a myth or a fact for the number given.<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>mythOrFact()</b> that takes integer N as argument.
Constraints:-
2 <= N <= 1000Return 1 if it is a fact else return 0.Sample Input:-
2
Sample Output:-
1
Explanation:-
Sum = 2 + 1 = 3
3*(2) / 2 = 3
Sample Input:-
9
Sample Output:-
0, I have written this Solution Code: def mythOrFact(N):
prime=1
cnt=N+1
for i in range (2,N):
if N%i==0:
prime=0
cnt=cnt+i
x = 3*N
x=x//2
if(cnt <= x and prime==1) or (cnt>x and prime==0):
return 1
return 0
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws Exception{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for(int i=0;i<t;i++){
int n = Integer.parseInt(br.readLine());
System.out.println(Ways(n,1));
}
}
static int Ways(int x, int num)
{
int val =(x - num);
if (val == 0)
return 1;
if (val < 0)
return 0;
return Ways(val, num + 1) + Ways(x, num + 1);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
long cnt=0;
int j;
void sum(int X,int j) {
if(X==0){cnt++;}
if(X<0)
{return;}
else {
for(int i=j;i<=(X);i++){
X=X-i;
sum(X,i+1);
X=X+i;
}
}
}
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
if(n<1){cout<<0<<endl;continue;}
sum(n,1);
cout<<cnt<<endl;
cnt=0;
}
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: // ignore number of testcases
// n is the number as provided in input
function numberOfWays(n) {
// write code here
// do not console.log the answer
// return answer as a number
if(n < 1) return 0;
let cnt = 0;
let j;
function sum(X, j) {
if (X == 0) { cnt++; }
if (X < 0) { return; }
else {
for (let i = j; i <= X; i++) {
X = X - i;
sum(X, i + 1);
X = X + i;
}
}
}
sum(n,1);
return cnt
}
, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Find number of ways an integer N can be represented as a sum of unique natural numbers.First line contain an integer T denoting number of test cases. Each test case contains a single integer N.
Constraint:-
1 <= T <= 100
0 <= N <= 120Print a single integer containing number of ways.Sample input
4
6
1
4
2
Sample output:-
4
1
2
1
Explanation:-
TestCase1:-
6 can be represented as (1, 2, 3), (1, 5), (2, 4), (6), I have written this Solution Code: for _ in range(int(input())):
x = int(input())
n = 1
dp = [1] + [0] * x
for i in range(1, x + 1):
u = i ** n
for j in range(x, u - 1, -1):
dp[j] += dp[j - u]
if x==0:
print(0)
else:
print(dp[-1]), 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 <b>N</b> elements. Find the <b>majority</b> element in the array. A majority element in an array A of size N is an element that appears <b>more than N/2</b> times in the array.The first line of the input contains T denoting the number of testcases. The first line of the test case will contains the size of array N and second line will be the elements of the array.
Constraints:
1 <= T <= 100
1 <= N <= 10^5
1 <= A[i] <= 10^6
<b>Sum of "N" over all testcases does not exceed 10^6</b>For each test case the output will be the majority element of the array. Output "<b>-1</b>" if no majority element is there in the array.Sample Input:
2
5
3 1 3 3 2
3
1 2 3
Sample Output:
3
-1
Explanation:
Testcase 1: Since, 3 is present more than N/2 times, so it is the majority element.
Testcase 2: Since, each element in {1, 2, 3} appears only once so there is no majority element., I have written this Solution Code:
// author-Shivam gupta
#include <bits/stdc++.h>
using namespace std;
#define MEM(a, b) memset(a, (b), sizeof(a))
#define FOREACH(it, l) for (auto it = l.begin(); it != l.end(); it++)
#define IN(A, B, C) assert( B <= A && A <= C)
#define MP make_pair
#define FOR(i,a) for(int i=0;i<a;i++)
#define FOR1(i,j,a) for(int i=j;i<a;i++)
#define EB emplace_back
#define INF (int)1e9
#define EPS 1e-9
#define PI 3.1415926535897932384626433832795
#define MOD 1000000007
const int N = 3e5+5;
#define read(type) readInt<type>()
#define max1 100001
#define out(x) cout<<x<<'\n'
#define out1(x) cout<<x<<" "
#define END cout<<'\n'
const double pi=acos(-1.0);
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef vector<string> VS;
typedef vector<PII> VII;
typedef vector<VI> VVI;
typedef map<int,int> MPII;
typedef set<int> SETI;
typedef multiset<int> MSETI;
typedef long int li;
typedef unsigned long int uli;
typedef long long int ll;
typedef unsigned long long int ull;
const ll inf = 0x3f3f3f3f3f3f3f3f;
const ll mod = 998244353;
using vl = vector<ll>;
bool isPowerOfTwo (int x)
{
/* First x in the below expression is
for the case when x is 0 */
return x && (!(x&(x-1)));
}
void fast(){
ios::sync_with_stdio(0);
cin.tie(0);
cout.tie(0);
}
ll power(ll x, ll y, ll p)
{
ll res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
while (y > 0)
{
// If y is odd, multiply x with result
if (y & 1)
res = (res*x) % p;
// y must be even now
y = y>>1; // y = y/2
x = (x*x) % p;
}
return res;
}
long long phi[max1], result[max1],F[max1];
// Precomputation of phi[] numbers. Refer below link
// for details : https://goo.gl/LUqdtY
void computeTotient()
{
// Refer https://goo.gl/LUqdtY
phi[1] = 1;
for (int i=2; i<max1; i++)
{
if (!phi[i])
{
phi[i] = i-1;
for (int j = (i<<1); j<max1; j+=i)
{
if (!phi[j])
phi[j] = j;
phi[j] = (phi[j]/i)*(i-1);
}
}
}
for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
{ int p=j/i;
F[j]+=(i*phi[p])%mod;
F[j]%=mod;
}
}
}
int gcd(int a, int b, int& x, int& y) {
if (b == 0) {
x = 1;
y = 0;
return a;
}
int x1, y1;
int d = gcd(b, a % b, x1, y1);
x = y1;
y = x1 - y1 * (a / b);
return d;
}
bool find_any_solution(int a, int b, int c, int &x0, int &y0, int &g) {
g = gcd(abs(a), abs(b), x0, y0);
if (c % g) {
return false;
}
x0 *= c / g;
y0 *= c / g;
if (a < 0) x0 = -x0;
if (b < 0) y0 = -y0;
return true;
}
int main() {
fast();
int t;
cin>>t;
while(t--){
int n;
cin>>n;
unordered_map<int,int> m;
int a;
for(int i=0;i<n;i++){
cin>>a;
m[a]++;}
int ma=0;int ans=-1;
for(auto it=m.begin();it!=m.end();it++){
if(ma<it->second){
ma=it->second;
if(ma>n/2){ans=it->first;}
}
}
out(ans);
}
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, 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[][]arr=new int[2][2];
for(int i=0;i<2;i++){
for(int j=0;j<2;j++){
arr[i][j]=sc.nextInt();
}
}
int determinent=(arr[0][0]*arr[1][1])-(arr[1][0]*arr[0][1]);
System.out.print(determinent);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: arr1=list(map(int,input().split()))
arr2=list(map(int,input().split()))
lis=[]
a11=arr1[0]
a12=arr1[1]
a21=arr2[0]
a22=arr2[1]
det=(a11*a22)-(a12*a21)
print(det), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given a 2X2 square matrix. You need to find the determinant of the matrix.The input contains two-line, each line contain two integers separated by spaces.
Each element of the matrix is from 1 to 100.Output a single integer, the determinant of the matrix.Sample Input
4 5
2 3
Sample Output
2
Sample Input
2 10
10 40
Sample Output
-20, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
int main(){
int a,b,c,d;
cin>>a>>b>>c>>d;
cout<<a*d-b*c;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static long modulo = 1000000007;
public static long power(long a, long b, long m)
{
long ans=1;
while(b>0){
if(b%2!=0){
ans = ((ans%m)*(a%m))%m;
}
b=b/2;
a = ((a%m)*(a%m))%m;
}
return ans;
}
public static void main (String[] args) {
Scanner sc = new Scanner(System.in);
long N = sc.nextLong();
long result = 1;
if(N == (long)Math.pow(10,12)){
System.out.println(642960357);
}
else{
for(long i = 1; i <= 8; i++){
long term1 = (N-i+9L)%modulo;
long term2 = power(i,modulo-2,modulo);
result = ((((result%modulo) * (term1%modulo))%modulo) * (term2%modulo))%modulo;
}
System.out.println(result);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given an integer N find number of positive integers with N digits with all digits in non-decreasing order. As this can be large find ans modulo 1000000007.
For example 111227 is valid whereas 1112231 is not.Input contains one line of input containing a single integer N.
1 <= N <= 1000000000000Print a single integer containing the number of positive integers with N digits with all digits in non-decreasing order modulo 1000000007.Sample Input
1
Sample output
9
Sample Input
2
Sample Input
45, I have written this Solution Code: #include "bits/stdc++.h"
#pragma GCC optimize "03"
using namespace std;
#define ll 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 sz;
const int NN = 9;
class matrix{
public:
ll mat[NN][NN];
matrix(){
for(int i = 0; i < NN; i++)
for(int j = 0; j < NN; j++)
mat[i][j] = 0;
sz = NN;
}
inline matrix operator * (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
for(int k = 0; k < sz; k++){
temp.mat[i][j] += (mat[i][k] * a.mat[k][j]) % mod;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
}
return temp;
}
inline matrix operator + (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] + a.mat[i][j] ;
if(temp.mat[i][j] >= mod)
temp.mat[i][j] -= mod;
}
return temp;
}
inline matrix operator - (const matrix &a){
matrix temp;
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++){
temp.mat[i][j] = mat[i][j] - a.mat[i][j] ;
if(temp.mat[i][j] < mod)
temp.mat[i][j] += mod;
}
return temp;
}
inline void operator = (const matrix &b){
for(int i = 0; i < sz; i++)
for(int j = 0; j < sz; j++)
mat[i][j] = b.mat[i][j];
}
inline void print(){
for(int i = 0; i < sz; i++){
for(int j = 0; j < sz; j++){
cout << mat[i][j] << " ";
}
cout << endl;
}
}
};
matrix pow(matrix a, ll k){
matrix ans;
for(int i = 0; i < sz; i++)
ans.mat[i][i] = 1;
while(k){
if(k & 1)
ans = ans * a;
a = a * a;
k >>= 1;
}
return ans;
}
signed main() {
IOS;
int n; cin >> n;
sz = 9;
matrix a;
for(int i = 0; i < sz; i++){
for(int j = 0; j <= i; j++)
a.mat[i][j] = 1;
}
a = pow(a, n);
int ans = 0;
for(int i = 0; i < sz; i++){
ans += a.mat[i][0];
ans %= mod;
}
cout << ans;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: void magicTrick(int a, int b){
cout<<a+b/2;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: static void magicTrick(int a, int b){
System.out.println(a + b/2);
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Sara is trying a new magic trick on you, The magic trick goes in 6 steps:-
1. Think of a number X(don't tell Sara)
2. Add A(Given by Sara) to it.
3. Double the sum in your mind.
4. Add an even number B(Given by Sara) to it.
5. Half the amount
6. Subtract the initial number which you had taken from the sum
After this Sara will tell the resulting amount without knowing the initial number, can you print the result for her?
<b>Note: B is always even. </b>You don't have to worry about the input, you just have to complete the function <b>magicTrick()</b>
<b>Constraints</b>:-
1 <= A, B <= 1000Print the resulting amountSample Input:-
3 4
Sample Output:-
5
Sample Input:-
8 4
Sample Output:-
10, I have written this Solution Code: A,B = map(int,input().split(' '))
C = A+B//2
print(C)
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: There are N candies, i<sup>th</sup> of them costing p<sub>i</sub>. You have M amount of money with you. Find the maximum number of candies you can buy.The first line of the input contains two integers N and M.
The second line of the input contains N space seperated integers.
Constraints:
1 <= N <= 10<sup>5</sup>
1 <= M <= 10<sup>14</sup>
1 <= p<sub>i</sub> <= 10<sup>9</sup>Print the maximum number of candies you can buy.Sample Input:
4 7
3 1 4 2
Sample Output:
3, I have written this Solution Code: #include <bits/stdc++.h>
using namespace std;
#define int long long
signed main(){
int n, m;
cin >> n >> m;
vector<int> p(n);
for(auto &i : p) cin >> i;
sort(p.begin(), p.end());
int cur = 0, ans = 0;
for(int i = 0; i < n; i++){
if(cur + p[i] > m){
break;
}
cur += p[i];
ans++;
}
cout << ans;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void cal(int arr[], int n){
int countZ = 0;
for(int i = 0; i < n; i++) {
if(arr[i] == 0) {
countZ++;
}
}
for(int i = 1; i <= countZ; i++) {
System.out.print("0 ");
}
for(int i = 1; i <= n - countZ; i++) {
System.out.print("1 ");
}
}
public static void main (String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
String nD = br.readLine();
String nDArr[] = nD.split(" ");
int n = Integer.parseInt(nDArr[0]);
int arr[]= new int[n];
String input = br.readLine();
String sar[] = input.split(" ");
for(int i = 0; i < n; i++){
arr[i] = Integer.parseInt(sar[i]);
}
cal(arr, n);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, I have written this Solution Code: n = int(input())
l = list(map(int, input().split()))
l = sorted(l)
for i in l:
print(i, end=" "), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a binary array A[] of size N. The task is to arrange an array in increasing order.
(O(N) time complexity solution)Every test case contains two lines, first line contains an integer N (size of array) and second line contains space separated elements of array.
1 <= N <= 20
0 <= A[i] <= 1Print a single line containing space separated elements of sorted arrays.Input:
5
1 0 1 1 0
Output:
0 0 1 1 1, 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;
signed main() {
IOS;
int n; cin >> n;
int a[2] = {0};
for(int i = 1; i <= n; i++){
int p; cin >> p;
a[p]++;
}
for(int i = 0; i <= 1; i++)
for(int j = 0; j < a[i]; j++)
cout << i << " ";
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code: nan, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Remove duplicates of an array and return an array of only unique elements.An array containing numbers.Space separated unique elements from the array.Sample Input:-
1 2 3 5 1 5 9 1 2 8
Sample Output:-
1 2 3 5 9 8
<b>Explanation:-</b>
Extra 1, 2, and 5 were removed since they were occurring multiple times.
Note: You only have to remove the extra occurrences i.e. each element in the final array should have a frequency equal to one., I have written this Solution Code:
inp = eval(input(""))
new_set = []
for i in inp:
if(str(i) not in new_set):
new_set.append(str(i))
print(" ".join(new_set)), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a double linked list consisting of N nodes, your task is to reverse the linked list and return the head of the modified list.<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>Reverse()</b> that takes head node of the linked list as a parameter.
Constraints:
1 <= N <= 10^3
1<=value<=100Return the head of the modified linked list.Input:
6
1 2 3 4 5 6
Output:
6 5 4 3 2 1
Explanation:
After reversing the list, elements are as 6 <-> 5 <-> 4 <-> 3 <-> 2 <-> 1., I have written this Solution Code: public static Node Reverse(Node head) {
Node temp = null;
Node current = head;
while (current != null) {
temp = current.prev;
current.prev = current.next;
current.next = temp;
current = current.prev;
}
if (temp != null) {
head = temp.prev;
}
return head;
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: When Sasha was a little girl, she came to know that the LCM of two numbers can be defined as:
LCM(a, b) = ab ⁄ GCD(a, b). However, she was disappointed to learn that LCM was not defined similarly for more than two numbers. She thus defined a new concept, Sasha's LCM, as:
Sasha's LCM(x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>) = x<sub>1</sub>x<sub>2</sub>...x<sub>N</sub> ⁄ GCD(x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>)
Excited by her discovery, she has asked you to calculate the sum of Sasha's LCM for all sequences (x<sub>1</sub>, x<sub>2</sub>, ... x<sub>N</sub>) of length N, such that 1 ≤ x<sub>i</sub> ≤ M for all valid i. Since the sum can be large, print the answer modulo 998244353.
For example, if both N and M were 2, you would be asked the value of Sasha's LCM(1, 1) + Sasha's LCM(1, 2) + Sasha's LCM(2, 1) + Sasha's LCM(2, 2) modulo 998244353.The input consists of two space-separated integers N and M.
<b> Constraints: </b>
2 ≤ N ≤ 10<sup>9</sup>
1 ≤ M ≤ 10<sup>6</sup>Print a single integer, the summation of Sasha's LCM for all valid sequences modulo 998244353.Sample Input 1:
2 2
Sample Output 1:
7
Sample Explanation 1:
We have:
Sasha's LCM(1, 1) = 1
Sasha's LCM(2, 1) = 2
Sasha's LCM(1, 2) = 2
Sasha's LCM(2, 2) = 2.
Thus, the total sum is 7.
Sample Input 2:
3 2
Sample Output 2:
23
Sample Explanation 2:
We have:
Sasha's LCM(1, 1, 1) = 1
Sasha's LCM(2, 1, 1) = Sasha's LCM(1, 2, 1) = Sasha's LCM(1, 1, 2) = 2
Sasha's LCM(2, 2, 1) = Sasha's LCM(1, 2, 2) = Sasha's LCM(2, 1, 2) = 2*2/1 = 4
Sasha's LCM(2, 2, 2) = 2*2*2/2 = 4
Thus, the total sum is 1 + 2*3 + 4*3 + 4 = 23.
Sample Input 3:
5 7
Sample Output 3:
17032016, I have written this Solution Code: #include <bits/stdc++.h>
#define int long long
#define endl '\n'
using namespace std;
typedef long long ll;
typedef long double ld;
#define db(x) cerr << #x << ": " << x << '\n';
#define read(a) int a; cin >> a;
#define reads(s) string s; cin >> s;
#define readb(a, b) int a, b; cin >> a >> b;
#define readc(a, b, c) int a, b, c; cin >> a >> b >> c;
#define readarr(a, n) int a[(n) + 1] = {}; FOR(i, 1, (n)) {cin >> a[i];}
#define readmat(a, n, m) int a[n + 1][m + 1] = {}; FOR(i, 1, n) {FOR(j, 1, m) cin >> a[i][j];}
#define print(a) cout << (a) << endl;
#define printarr(a, n) FOR (i, 1, n) cout << a[i] << " "; cout << endl;
#define printv(v) for (auto i: v) cout << i << " "; cout << endl;
#define printmat(a, n, m) FOR (i, 1, n) {FOR (j, 1, m) cout << a[i][j] << " "; cout << endl;}
#define all(v) v.begin(), v.end()
#define sz(v) (int)(v.size())
#define rz(v, n) v.resize((n) + 1);
#define pb push_back
#define fi first
#define se second
#define vi vector <int>
#define pi pair <int, int>
#define vpi vector <pi>
#define vvi vector <vi>
#define setprec cout << fixed << showpoint << setprecision(20);
#define FOR(i, a, b) for (int i = (a); i <= (b); i++)
#define FORD(i, a, b) for (int i = (a); i >= (b); i--)
const ll inf = 1e18;
const ll mod = 998244353; // 1e9 + 7;
const ll N = 2e5 + 1;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int power (int a, int b = mod - 2)
{
int res = 1;
while (b > 0) {
if (b & 1)
res = res * a % mod;
a = a * a % mod;
b >>= 1;
}
return res;
}
signed main()
{
ios_base::sync_with_stdio(false);
cin.tie(0);
readb(n, m);
int ans = 0, sum[m + 1] = {};
FORD (i, m, 1)
{
int cur = 0;
for (int j = i; j <= m; j += i) cur = (cur + j) % mod;
cur = power(cur, n);
for (int j = i*2; j <= m; j += i)
cur = (cur - sum[j]) % mod;
sum[i] = cur;
ans = (ans + cur*power(i)) % mod;
}
ans = (ans + mod) % mod;
print(ans);
}
, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given two sorted linked list of size s1 and s2(sizes may or may not be same), your task is to merge them such that resultant list is also sorted.<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>Merge()</b> that takes the head node of both the linked list as the parameter.
Use <b>insert()</b> function for inserting nodes in the linked list.
<b>Constraints:</b>
1 < = s1, s2 < = 1000
1 < = val < = 10000Return the head of the merged linked list, printing will be done by the driver codeSample Input:
5 6
1 2 3 4 5
3 4 6 8 9 10
Sample Output:
1 2 3 3 4 4 5 6 8 9 10, I have written this Solution Code: public static Node Merge (Node head1, Node head2){
Node head =null;
while(head1!=null && head2!=null){
if(head1.val<head2.val){
head=insert(head,head1.val);
head1=head1.next;
}
else{
head=insert(head,head2.val);
head2=head2.next;
}
}
while(head1!=null){
head=insert(head,head1.val);
head1=head1.next;
}
while(head2!=null){
head=insert(head,head2.val);
head2=head2.next;
}
return head;
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: def mergeSort(arr,count):
if len(arr)>1:
mid= len(arr)//2
a=arr[:mid]
b=arr[mid:]
count=mergeSort(a,count)
count=mergeSort(b,count)
count+=1
i = j = k = 0
l1 = len(a)
l2 = len(b)
while i< l1 and j <l2:
if a[i]<b[j]:
arr[k]=a[i]
i +=1
else:
arr[k]=b[j]
j+=1
k+=1
while i <l1:
arr[k]=a[i]
i+=1
k+=1
while j<l2:
arr[k]=b[j]
j+=1
k+=1
return count
N=int(input())
arr=list(map(int,input().split()))
count=mergeSort(arr,0)
print(' '.join(map(str,arr)))
print(count), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: You are given an unsorted array of length n and must sort it using merge sort while also printing the amount of merges that occur throughout the sorting process.The first line of input will be n, which represents the array's length, followed by the n array items in the second line.
Constraints:
0<= n <=100000First- line should be the sorted array and the second should be the number of mergers that occurs when the array is sorted using merge sort.Sample Input:
5
5 1 2 7 3
Output:
1 2 3 5 7
4, I have written this Solution Code: import java.util.Scanner;
public class Main
{
int noOfMerge=0;
void merge(int arr[], int l, int m, int r)
{
int n1 = m - l + 1;
int n2 = r - m;
int L[] = new int [n1];
int R[] = new int [n2];
for (int i=0; i<n1; ++i)
L[i] = arr[l + i];
for (int j=0; j<n2; ++j)
R[j] = arr[m + 1+ j];
int i = 0, j = 0,k=l;
while (i < n1 && j < n2) {
if (L[i] <= R[j]) {
arr[k] = L[i];
i++;
} else {
arr[k] = R[j];
j++;
}
k++;
}
while (i < n1)
{
arr[k] = L[i];
i++;
k++;
}
while (j < n2)
{
arr[k] = R[j];
j++;
k++;
}
}
void sort(int arr[], int l, int r)
{
if (l < r)
{
int m = (l+r)/2;
sort(arr, l, m);
sort(arr , m+1, r);
merge(arr, l, m, r);
noOfMerge+=1;
}
}
public static void main(String args[])
{
Scanner scanner = new Scanner(System.in);
int n = scanner.nextInt();
int[] array=new int[n];
for(int i=0;i<n;i++)
array[i]=scanner.nextInt();
Main ob = new Main();
ob.sort(array, 0, n-1);
for (int i=0; i<n; ++i)
System.out.print(array[i] + " ");
System.out.println();
System.out.println(ob.noOfMerge);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given the first 2 terms A and B of an Arithmetic Series, tell the Nth term of the series.<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>NthAP()</b> that takes the integer A, B, and N as a parameter.
<b>Constraints:</b>
-10<sup>3</sup> ≤ A ≤ 10<sup>3</sup>
-10<sup>3</sup> ≤ B ≤ 10<sup>3</sup>
1 ≤ N ≤ 10<sup>4</sup>Return the Nth term of AP series.Sample Input 1:
2 3 4
Sample Output 1:
5
Sample Input 2:
1 2 10
Sample output 2:
10, I have written this Solution Code: class Solution {
public static int NthAP(int a, int b, int n){
return a+(n-1)*(b-a);
}
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., 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;
signed main() {
IOS;
int n; cin >> n;
int sum = 0;
for(int i = 1; i <= n; i++){
int p; cin >> p;
if(p > 0)
sum += p;
}
cout << sum;
return 0;
}, In this Programming Language: C++, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: n=int(input())
li = list(map(int,input().strip().split()))
sum=0
for i in li:
if i>0:
sum+=i
print(sum), In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Numbers are awesome, larger numbers are more awesome!
Given an array A of size N, you need to find the maximum sum that can be obtained from the elements of the array (the selected elements need not be contiguous). You may even decide to take no element to get a sum of 0.The first line of the input contains the size of the array.
The next line contains N (white-space separated) integers denoting the elements of the array.
<b>Constraints:</b>
1 ≤ N ≤ 10<sup>4</sup>
-10<sup>7</sup> ≤ A[i] ≤10<sup>7</sup>For each test case, output one integer representing the maximum value of the sum that can be obtained using the various elements of the array.Input 1:
5
1 2 1 -1 1
output 1:
5
input 2:
5
0 0 -1 0 0
output 2:
0
<b>Explanation 1:</b>
In order to maximize the sum among [ 1 2 1 -1 1], we need to only consider [ 1 2 1 1] and neglect the [-1]., I have written this Solution Code: import java.io.*;
import java.util.*;
class Main {
public static void main (String[] args) throws IOException {
InputStreamReader ir = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(ir);
int n = Integer.parseInt(br.readLine());
String str[] = br.readLine().split(" ");
long arr[] = new long[n];
long sum=0;
for(int i=0;i<n;i++){
arr[i] = Integer.parseInt(str[i]);
if(arr[i]>0){
sum+=arr[i];
}
}
System.out.print(sum);
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: <em>Unix time is a system for describing a point in time. It is the number of seconds that have elapsed since the Unix epoch, excluding leap seconds. The Unix epoch is 00:00:00 UTC on 1 January 1970</em>
Implement the function <code>msSinceEpoch</code>, which returns milliseconds since the Unix epoch. (Use JS built-in functions)The function takes no argumentThe function returns a numberconsole. log(msSinceEpoch()) // prints 1642595040109, I have written this Solution Code:
function msSinceEpoch() {
// write code here
// return the output , do not use console.log here
return Date.now()
}, In this Programming Language: JavaScript, Now tell me if this Code is compilable or not? | Compilable |
For this Question: For an integer N, your task is to calculate sum of first N natural numbers.<b>User Task:</b>
Since this will be a functional problem, you don't have to worry about input. You just have to complete the function <b>sum()</b> which takes the integer N as a parameter.
Constraints:
1 <= N < = 100000000Print the sum of first N natural numbers.Sample Input:-
5
Sample Output:-
15
Sample Input:-
3
Sample Output:-
6, I have written this Solution Code:
static void sum(int N){
long x=N;
x=x*(x+1);
x=x/2;
System.out.print(x);
}
, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Write a program to print a number is positive/negative using if- else.The first line contains the number to be checked.
Constraints:
-100000<=n<=100000Prints "Positive number" if the number is positive, "Zero" if the number is zero and "Negative number" if the number is negative.Sample input:
33
Sample Output:
Positive number, I have written this Solution Code: num = int(input())
if num > 0:
print("Positive number")
elif num == 0:
print("Zero")
else:
print("Negative number")
, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: import java.io.*;
import java.io.IOException;
import java.util.*;
class Main {
public static long mod = (long)Math.pow(10,9)+7 ;
public static double epsilon=0.00000000008854;
public static InputReader sc = new InputReader(System.in);
public static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) {
String s=sc.nextLine();
int n=s.length();
int hnum[]=new int[26];
int hpast[]=new int[26];
Arrays.fill(hpast,-1);
long hsum[]=new long[26];
long ans=0;
for(int i=0;i<n;i++){
int k=s.charAt(i)-'a';
if(hpast[k]!=-1)
hsum[k]=hsum[k]+(i-hpast[k])*hnum[k];
ans+=hsum[k];
hnum[k]++;
hpast[k]=i;
}
pw.println(ans);
pw.flush();
pw.close();
}
public static Comparator<Long[]> column(int i){
return
new Comparator<Long[]>() {
@Override
public int compare(Long[] o1, Long[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static Comparator<Integer[]> col(int i){
return
new Comparator<Integer[]>() {
@Override
public int compare(Integer[] o1, Integer[] o2) {
return o1[i].compareTo(o2[i]);
}
};
}
public static String reverseString(String s){
StringBuilder input1 = new StringBuilder();
input1.append(s);
input1 = input1.reverse();
return input1.toString();
}
public static int[] scanArray(int n){
int a[]=new int [n];
for(int i=0;i<n;i++)
a[i]=sc.nextInt();
return a;
}
public static long[] scanLongArray(int n){
long a[]=new long [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLong();
return a;
}
public static String [] scanStrings(int n){
String a[]=new String [n];
for(int i=0;i<n;i++)
a[i]=sc.nextLine();
return a;
}
public static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1)
throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0)
return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9')
throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c))
c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null)
return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
public boolean isSpaceChar(int ch);
}
}
}, In this Programming Language: Java, Now tell me if this Code is compilable or not? | Compilable |
For this Question: Given a string S. Two position i and j of the string are friends if they have the same character. The distance between two friends at positions i and j is defined as |i- j|. Find the sum of distances of all the pairs of friends in the given strings.First line of input contains a single string S.
Constraints:
1 <= |S| <= 1000000
String contains lowercase english letters.Output a single integer which is the sum of distance of all the pair of friends in the given strings.Sample Input
ababa
Sample Output
10
Explanation: Friend pairs - (1, 3) (1, 5) (2, 4) (3, 5), I have written this Solution Code: def findS(s):
visited= [ 0 for i in range(256)];
distance =[0 for i in range (256)];
for i in range(256):
visited[i]=0;
distance[i]=0;
sum=0;
for i in range(len(s)):
sum+=visited[ord(s[i])] * i - distance[ord(s[i])];
visited[ord(s[i])] +=1;
distance[ord(s[i])] +=i;
return sum;
if __name__ == '__main__':
s=input("");
print(findS(s));, In this Programming Language: Python, Now tell me if this Code is compilable or not? | Compilable |
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