```
// A divide and conquer solution to find count of zeroes in an array
// where all 1s are present before all 0s
#include <stdio.h>
/* if 0 is present in arr[] then returns the index of FIRST occurrence
of 0 in arr[low..high], otherwise returns -1 */
int firstZero(int arr[], int low, int high)
{
if (high >= low)
{
// Check if mid element is first 0
int mid = low + (high - low)/2;
if (( mid == 0 || arr[mid-1] == 1) && arr[mid] == 0)
return mid;
if (arr[mid] == 1) // If mid element is not 0
return firstZero(arr, (mid + 1), high);
else // If mid element is 0, but not first 0
return firstZero(arr, low, (mid -1));
}
return -1;
}
// A wrapper over recursive function firstZero()
int countOnes(int arr[], int n)
{
// Find index of first zero in given array
int first = firstZero(arr, 0, n-1);
// If 0 is not present at all, return 0
if (first == -1)
return 0;
return (n - first);
}
/* Driver program to check above functions */
int main()
{
int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};
int n = sizeof(arr)/sizeof(arr[0]);
printf("Count of zeroes is %d", countOnes(arr, n));
return 0;
}
Output:
Count of zeroes is 5
```