Minimum Toggles to Partition Binary Array in C++

Problem statement

Given an array of n integers containing only 0 and 1. Find the minimum toggles (switch from 0 to 1 or vice-versa) required such the array the array become partitioned, i.e., it has first 0s then 1s.

Example

If arr[] = {1, 0, 0, 1, 1, 1, 0} then 2 toggle is required i.e. toggle first one and last zero.

Algorithm

• If we observe the question, then we will find that there will definitely exist a point from 0 to n-1 where all elements left to that point should contains all 0’s and right to point should contains all 1’s
• Those indices which don’t obey this law will have to be removed. The idea is to count all 0s from left to right.

Example

 Live Demo

#include <bits/stdc++.h>
using namespace std;
int getMinToggles(int *arr, int n) {
   int zeroCnt[n + 1] = {0};
   for (int i = 1; i <= n; ++i) {
      if (arr[i - 1] == 0) {
         zeroCnt[i] = zeroCnt[i - 1] + 1;
      } else {
         zeroCnt[i] = zeroCnt[i - 1];
      }
   }
   int result = n;
   for (int i = 1; i <= n; ++i) {
      result = min(result, i - zeroCnt[i] + zeroCnt[n] - zeroCnt[i]);
   }
   return result;
}
int main() {
   int arr[] = {1, 0, 0, 1, 1, 1, 0};
   int n = sizeof(arr) / sizeof(arr[0]);
   cout << "Minimum toggles = " << getMinToggles(arr, n) << endl;
   return 0;
}

When you compile and execute above program. It generates following output −

Output

Minimum toggles = 2