# Find the number of zeroes

Given an array of 1s and 0s which has all 1s first followed by all 0s. Find the number of 0s. Count the number of zeroes in the given array.

Examples:

```Input: arr[] = {1, 1, 1, 1, 0, 0}
Output: 2

Input: arr[] = {1, 0, 0, 0, 0}
Output: 4

Input: arr[] = {0, 0, 0}
Output: 3

Input: arr[] = {1, 1, 1, 1}
Output: 0
```

## Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

A simple solution is to traverse the input array. As soon as we find a 0, we return n – index of first 0. Here n is number of elements in input array. Time complexity of this solution would be O(n).

Since the input array is sorted, we can use Binary Search to find the first occurrence of 0. Once we have index of first element, we can return count as n – index of first zero.

## C

```// 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 countZeroes(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", countZeroes(arr, n));
return 0;
}
```

## Java

```// A divide and conquer solution to find count of zeroes in an array
// where all 1s are present before all 0s

class CountZeros
{
/* 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 countZeroes(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 test above functions
public static void main(String[] args)
{
CountZeros count = new CountZeros();
int arr[] = {1, 1, 1, 0, 0, 0, 0, 0};
int n = arr.length;
System.out.println("Count of zeroes is " + count.countZeroes(arr, n));
}
}

```

Output:
`Count of zeroes is 5 `

Time Complexity: O(Logn) where n is number of elements in arr[].

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