Javascript Program to Count 1's in a sorted binary array

Last Updated : 16 Sep, 2024

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 traverse the array linearly. 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 the 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 the above idea.

JavaScript

// 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;
console.log("Count of 1's in given array is " + countOnes(arr, 0, n - 1));

Output

Count of 1's in given array is 4

Complexity Analysis:

Time complexity: O(Logn).
Space complexity: o(log n) (function call stack)

The same approach with iterative solution would be

JavaScript

/* 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;
console.log("Count of 1's in given array is " + countOnes(arr, n));
// This code is contributed by unknown2108

Output

Count of 1's in given array is 4

Complexity Analysis:

Time complexity: O(Logn)
Space complexity: O(1)

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

kartik

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Binary Search

Javascript Program to Count 1's in a sorted binary array

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