# Count 1’s in a sorted binary array

Given a binary array sorted in non-increasing order, count the number of 1’s in it.

Examples:

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

Input: arr[] = {1, 1, 1, 1, 1, 1, 1}
Output: 7

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

A simple solution is to linearly traverse the array. The time complexity of the simple solution is O(n). We can use Binary Search to find count in O(Logn) time. The idea is to look for last occurrence of 1 using Binary Search. Once we find the index last occurrence, we return index + 1 as count.

The following is C++ implementation of above idea.

```// C++ program to count one's in a boolean array

#include <iostream>

using namespace std;

/* Returns counts of 1's in arr[low..high].  The array is

assumed to be sorted in non-increasing order */

int countOnes(bool arr[], int low, int high)

{

if (high >= low)

{

// get the middle index

int mid = low + (high - low)/2;

// check if the element at middle index is last 1

if ( (mid == high || arr[mid+1] == 0) && (arr[mid] == 1))

return mid+1;

// If element is not last 1, recur for right side

if (arr[mid] == 1)

return countOnes(arr, (mid + 1), high);

// else recur for left side

return countOnes(arr, low, (mid -1));

}

return 0;

}

/* Driver program to test above functions */

int main()

{

bool arr[] = {1, 1, 1, 1, 0, 0, 0};

int n = sizeof(arr)/sizeof(arr);

cout << "Count of 1's in given array is " << countOnes(arr, 0, n-1);

return 0;

}```

Output:

`Count of 1's in given array is 4`

Time complexity of the above solution is O(Logn)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above

Source : GeeksforGeeks