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JS binary search Code Example
binary search in js
function binarySearch(arr, val) {
  let start = 0;
  let end = arr.length - 1;

  while (start <= end) {
    let mid = Math.floor((start + end) / 2);

    if (arr[mid] === val) {
      return mid;
    }

    if (val < arr[mid]) {
      end = mid - 1;
    } else {
      start = mid + 1;
    }
  }
  return -1;
}
binary search javascript
const arr = [1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30];


const binarySearch = (arr, x , start=0, end=arr.length) => {
  if(end < start) return -1;
  
  let mid = Math.floor((start + end) / 2);
  
  if(arr[mid] === x) return mid;

  if(arr[mid] < x) return binarySearch(arr, x, mid+1, end);
  
  else return binarySearch(arr, x , start, mid-1);
}

console.log(arr[binarySearch(arr,8)]);
JS binary search

// JavaScript program to implement recursive Binary Search
  
// A recursive binary search function. It returns
// location of x in given array arr[l..r] is present,
// otherwise -1
function binarySearch(arr, l, r, x){
    if (r >= l) {
        let mid = l + Math.floor((r - l) / 2);
  
        // If the element is present at the middle
        // itself
        if (arr[mid] == x)
            return mid;
  
        // If element is smaller than mid, then
        // it can only be present in left subarray
        if (arr[mid] > x)
            return binarySearch(arr, l, mid - 1, x);
  
        // Else the element can only be present
        // in right subarray
        return binarySearch(arr, mid + 1, r, x);
    }
  
    // We reach here when element is not
    // present in array
    return -1;
}
  
let arr = [ 2, 3, 4, 10, 40 ];
let x = 10;
let n = arr.length
let result = binarySearch(arr, 0, n - 1, x);
(result == -1) ? document.write( "Element is not present in array")                   : document.write("Element is present at index " +result);




Javascript

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Code Example
Category:
Coding
Sub Category:
Code Example
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