Binary Search In JavaScript
Last Updated :
04 Dec, 2023
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:
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.
Example: This example shows the use of the above-explained approach.
javascript
let recursiveFunction = function (arr, x, start, end) {
if (start > end) return false ;
let mid = Math.floor((start + end) / 2);
if (arr[mid] === x) return true ;
if (arr[mid] > x)
return recursiveFunction(arr, x, start, mid - 1);
else
return recursiveFunction(arr, x, mid + 1, end);
}
let arr = [1, 3, 5, 7, 8, 9];
let x = 5;
if (recursiveFunction(arr, x, 0, arr.length - 1)) {
console.log( "Element found!" );
}
else { console.log( "Element not found!" ); }
x = 6;
if (recursiveFunction(arr, x, 0, arr.length - 1)) {
console.log( "Element found!" );
}
else { console.log( "Element not found!" ); }
|
Output
Element found!
Element not found!
Time Complexity: O(logN)
Auxiliary Space: O(1)
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
let iterativeFunction = function (arr, x) {
let start = 0, end = arr.length - 1;
while (start <= end) {
let mid = Math.floor((start + end) / 2);
if (arr[mid] === x) return true ;
else if (arr[mid] < x)
start = mid + 1;
else
end = mid - 1;
}
return false ;
}
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)
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