HackerEarth Average Subarray problem solution

In this HackerEarth Average Subarray problem solution, You are given an binary array A of N integers. You are also given an integer K and you need to ensure that no subarray of size greater than or equal to K has average 1.

For this you can perform the below operation:

Choose any index and flip the bit.

Find the minimum number of operations to acheive the above condition.

HackerEarth Average Subarray problem solution.

#include<bits/stdc++.h>
using namespace std;
#define FIO ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define mod 1000000007
#define endl "n"
#define test ll t; cin>>t; while(t--)
typedef long long int ll;
int main() {
    FIO;
    test
    {
       ll n,q; cin>>n>>q;
       vector<ll>x(n),r(n);
       for(auto &it:x) cin>>it;
       for(auto &it:r) cin>>it;
       vector<ll>ans(4*n+5,0);
       for(int i=0;i<n;i++){
           int left=x[i]-r[i]+2*n;
           int right=x[i]+r[i]+2*n+1;
           if(x[i]>0){
               left=max(left,2*n);
           }
           else{
               right=max(right,2*n);
           }
           ans[left]++;
           ans[right]--;
       }
       for(int i=1;i<4*n+5;i++){
           ans[i]+=ans[i-1];
       }
       while(q--){
           int inp; cin>>inp;
           inp+=2*n;
           cout<<ans[inp]<<endl;
       }
    }
    return 0;
}

Second solution

t = int(input())

while t > 0:
    t -= 1
    n, k = map(int, input().split())
    cnt = ans = 0
    for x in input().split():
        if x == '1':
            cnt += 1
            if cnt == k:
                ans += 1
                cnt = 0
        else:
            cnt = 0
    print(ans)

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