# 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)`