binarySearch.js

// LeetCode #704. Binary Search
/*
Given an array of integers nums which is sorted in ascending order, and an integer target, write a function to search target in nums. If target exists, then return its index. Otherwise, return -1.

You must write an algorithm with O(log n) runtime complexity.

 

Example 1:

Input: nums = [-1,0,3,5,9,12], target = 9
Output: 4
Explanation: 9 exists in nums and its index is 4

Example 2:

Input: nums = [-1,0,3,5,9,12], target = 2
Output: -1
Explanation: 2 does not exist in nums so return -1

 

Constraints:

    1 <= nums.length <= 104
    -104 < nums[i], target < 104
    All the integers in nums are unique.
    nums is sorted in ascending order.

*/

// Solution
function binarySearch(nums, target) {
    let left = 0
    let right = nums.length - 1

    while (left < right) {
        let mid = left + Math.floor((right-left + 1) / 2)
        if (target < nums[mid]) {
            right = mid - 1
        } else {
            left = mid
        }
    }

    return nums[left] === target ? left : -1
}

// Refactored
// const binarySearch = (nums, target) => {
//     let left = 0
//     let right = nums.length - 1

//     while (left < right) {
//         let mid = left + Math.floor((right - left + 1) / 2)
//         if (target < nums[mid]) {
//             right = mid - 1
//         } else {
//             left = mid
//         }
//     }

//     return nums[left] === target ? left : -1
// }