Minimum number of operations on an array to make all elements 0 using C++.

Problem statement

Given an array of size N and each element is either 1 or 0. The task is to calculated the minimum number of operations to be performed to convert all elements to zero. One can perform below operations −

If an element is 1, You can change its value equal to 0 then −

If the next consecutive element is 1, it will automatically get converted to 0
If the next consecutive element is already 0, nothing will happen.

If arr[] = {1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1} then 4
operation are required to make all elements zero

Algorithm

1.If the current element is 1 then increment the count and search for the next 0 as all consecutive 1’s will be automatically converted to 0.
2. Return final count

Example

#include <iostream>
#define SIZE(arr) (sizeof(arr) / sizeof(arr[0]))
using namespace std;
int performMinOperation(int *arr, int n){
   int i, cnt = 0;
   for (i = 0; i < n; ++i) {
      if (arr[i] == 1) {
         int j;
         for (j = i + 1; j < n; ++j) {
            if (arr[j] == 0) {
               break;
            }
         }
         i = j - 1;
         ++cnt;
      }
   }
   return cnt;
}
int main(){
   int arr[] = {1, 1, 0, 0, 1, 1, 1, 1, 1, 0, 0, 1, 0, 0, 1};
   cout << "Minimum required operations = " << performMinOperation(arr, SIZE(arr)) << endl;
   return 0;
}