Maximize number of 0s by flipping a subarray in C++

Problem statement

Given a binary array, find the maximum number of zeros in an array with one flip of a subarray allowed. A flip operation switches all 0s to 1s and 1s to 0s

If arr1= {1, 1, 0, 0, 0, 0, 0}

If we flip first 2 1’s to 0’s, then we can get subarray of size 7 as follows −

{0, 0, 0, 0, 0, 0, 0}

Algorithm

1. Consider all subarrays and find a subarray with maximum value of (count of 1s) – (count of 0s)
2. Considers this value be maxDiff. Finally return count of zeros in original array + maxDiff.

Example

Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMaxSubArray(int *arr, int n){
   int maxDiff = 0;
   int zeroCnt = 0;
   for (int i = 0; i < n; ++i) {
      if (arr[i] == 0) {
         ++zeroCnt;
      }
      int cnt0 = 0;
      int cnt1 = 0;
      for (int j = i; j < n; ++j) {
         if (arr[j] == 1) {
            ++cnt1;
         }
         else {
            ++cnt0;
         }
         maxDiff = max(maxDiff, cnt1 - cnt0);
      }
   }
   return zeroCnt + maxDiff;
}
int main(){
   int arr[] = {1, 1, 0, 0, 0, 0, 0};
   int n = sizeof(arr) / sizeof(arr[0]);
   cout << "Maximum subarray size = " << getMaxSubArray(arr, n) << endl;
   return 0;
}