leetcode/apps/docs/content/(array)/704-binary-search.mdx

---
title: '704. Binary Search'
description: 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
sidebar:
  label: 'Binary Search'
  badge: 'Easy'
---

<Badge variant="accent">Binary Search</Badge>

::::warning
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 <= 10^4`
- `-10^4 < nums[i], target < 10^4`
- All the integers in `nums` are unique.
- `nums` is sorted in ascending order.

## Approach

```mermaid
flowchart TD
  S(["search(nums, target)"]) --> I["left = 0, right = n - 1"]
  I --> W{"left <= right?"}
  W -- no --> E(["return -1"])
  W -- yes --> M["mid = left + floor((right - left) / 2)"]
  M --> Q{"nums[mid] vs target"}
  Q -- equal --> R(["return mid"])
  Q -- "nums[mid] < target" --> L["left = mid + 1"]
  L --> W
  Q -- "nums[mid] > target" --> H["right = mid - 1"]
  H --> W
```

## Solution

```js
/**
 * @param {number[]} nums
 * @param {number} target
 * @return {number}
 */
var search = function(nums, target) {
    let left = 0;
    let right = nums.length - 1;

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

## Explanation

<YouTube id="s4DPM8ct1pI" title="Binary Search - Leetcode 704 - Python" />