Javascript Program to Count 1’s in a sorted binary array
Last Updated :
16 Sep, 2024
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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 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
/* 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!
