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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
#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;
}Output
When you compile and execute the above program. It generates the following output−
Maximum subarray size = 7
Updated on: 24-Dec-2019
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