Minimum toggles to partition a binary array so that it has first 0s then 1s 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 −