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# 704. Binary Search

🟩 Easy

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

## Solution

My Solution

```go
func search(nums []int, target int) int {
    l, r := 0, len(nums)-1
    for l <= r {
        mid := l+(r-l) / 2
        if nums[mid] == target {
            return mid
        } else if nums[mid] < target {
            l = mid+1
        } else {
            r = mid - 1
        }
    }

    return -1
}
```

Optimal Solution (Binary Search)

```go
func search(nums []int, target int) int {
    left, right := 0, len(nums)-1
    
    for left <= right {
        mid := left + (right-left)/2  // Prevents integer overflow
        
        switch {
        case nums[mid] == target:
            return mid
        case nums[mid] < target:
            left = mid + 1
        default:
            right = mid - 1
        }
    }
    
    return -1
}
```

### Approach

Binary Search is an efficient algorithm for searching in a sorted array:

1. Initialize pointers:
   * Left pointer at start (index 0)
   * Right pointer at end (index n-1)
2. While left ≤ right:
   * Calculate middle point: mid = left + (right-left)/2
   * Using (right-left)/2 instead of (left+right)/2 prevents integer overflow
3. Compare middle element with target:
   * If equal: found target, return index
   * If target > middle: search right half (left = mid+1)
   * If target < middle: search left half (right = mid-1)
4. If loop ends without finding target:
   * Return -1 (target not found)

### Complexity Analysis

#### Time Complexity: O(log n)

* Each iteration eliminates half of the remaining elements
* The search space is halved in each step
* Takes log₂(n) steps to reduce n elements to 1

#### Space Complexity: O(1)

* Only uses three variables regardless of input size:
  * left - left pointer
  * right - right pointer
  * mid - middle index

### Why it works

* Takes advantage of the sorted nature of the array
* Efficiently eliminates half of the remaining elements in each step
* Using (right-left)/2 prevents integer overflow that could occur with (left+right)/2
* The loop condition left <= right ensures we check all possible elements

![result](/files/TGPUFOcUOObqtZAlWxxC)

Leetcode: [link](https://leetcode.com/problems/binary-search/description/)