Javascript Program to 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 the implementation of above idea. 

Javascript

<script>

// Javascript program to count one's in a boolean array

/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */
function countOnes( arr, low, high)
{
  if (high >= low)
  {
    // get the middle index
    let mid = Math.trunc(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

   let arr = [ 1, 1, 1, 1, 0, 0, 0 ];
   let n = arr.length;
   document.write("Count of 1's in given array is " +
                    countOnes(arr, 0, n-1));

</script>

Output

Count of 1's in given array is 4

Time complexity of the above solution is O(Logn)

Space complexity o(log n) (function call stack)

The same approach with iterative solution would be

Javascript

<script>
/* Returns counts of 1's in arr[low..high].  The array is
   assumed to be sorted in non-increasing order */

function countOnes(arr,n)
{
    let ans;
    let low = 0, high = n - 1;
    while (low <= high) { // get the middle index
        let mid = Math.floor((low + high) / 2);

        // else recur for left side
        if (arr[mid] < 1)
            high = mid - 1;
        // If element is not last 1, recur for right side
        else if (arr[mid] > 1)
            low = mid + 1;
        else
        // check if the element at middle index is last 1
        {
            if (mid == n - 1 || arr[mid + 1] != 1)
                return mid + 1;
            else
                low = mid + 1;
        }
    }
}

let arr=[ 1, 1, 1, 1, 0, 0, 0];
let n = arr.length;
document.write( "Count of 1's in given array is "+ countOnes(arr, n));


// This code is contributed by unknown2108
</script>

Output

Count of 1's in given array is 4

Time complexity of the above solution is O(Logn)

Space complexity is O(1)

Please refer complete article on Count 1’s in a sorted binary array for more details!