Iterative Approach

In this iterative approach, instead of recursion, we use a while loop, and the loop runs until it hits the base condition, i.e. start becomes greater than end. 

Example: This example shows the use of the above-explained approach.

javascript




// Iterative function to implement Binary Search
let iterativeFunction = function (arr, x) {
 
    let start = 0, end = arr.length - 1;
 
    // Iterate while start not meets end
    while (start <= end) {
 
        // Find the mid index
        let mid = Math.floor((start + end) / 2);
 
        // If element is present at
        // mid, return True
        if (arr[mid] === x) return true;
 
        // Else look in left or
        // right half accordingly
        else if (arr[mid] < x)
            start = mid + 1;
        else
            end = mid - 1;
    }
 
    return false;
}
 
// Driver code
let arr = [1, 3, 5, 7, 8, 9];
let x = 5;
 
if (iterativeFunction(arr, x, 0, arr.length - 1)) {
    console.log("Element found!");
}
else {
    console.log("Element not found!");
}
 
x = 8;
 
if (iterativeFunction(arr, x, 0, arr.length - 1)) {
    console.log("Element found!");
}
else {
    console.log("Element not found!");
}


Output

Element found!
Element found!

Time Complexity: O(logN).

Auxiliary Space: O(1) 



Binary Search In JavaScript

Binary Search is a searching technique that works on the Divide and Conquer approach. It is used to search for any element in a sorted array. Compared with linear, binary search is much faster with a Time Complexity of O(logN), whereas linear search works in O(N) time complexity

Examples:  

Input : arr[] = {1, 3, 5, 7, 8, 9}, x = 5
Output : Element found!

Input : arr[] = {1, 3, 5, 7, 8, 9}, x = 6
Output : Element not found!

Note: Assuming the array is sorted.

These are the following ways to do Binary Search in JavaScript: 

Table of Content

  • Recursive Approach
  • Iterative Approach

Similar Reads

Recursive Approach:

BASE CONDITION: If the starting index is greater than the ending index return false. Compute the middle index. Compare the middle element with the number x. If equal return true. If greater, call the same function with ending index = middle-1 and repeat step 1. If smaller, call the same function with starting index = middle+1 and repeat step 1....

Iterative Approach:

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