033. Search in Rotated Sorted Array

At a Glance

• Topic: Array
• Pattern: Analyze Pattern
• Difficulty: Medium
• LeetCode: 033

Problem Statement

There is an integer array nums sorted in ascending order (with distinct values).

Prior to being passed to your function, nums is possibly left rotated at an unknown index k (1 <= k < nums.length) such that the resulting array is [nums[k], nums[k+1], ..., nums[n-1], nums[0], nums[1], ..., nums[k-1]] (0-indexed). For example, [0,1,2,4,5,6,7] might be left rotated by 3 indices and become [4,5,6,7,0,1,2].

Given the array nums after the possible rotation and an integer target, return the index of target if it is in nums, or -1 if it is not in nums.

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

Example 1: Input: nums = [4,5,6,7,0,1,2], target = 0 Output: 4 Example 2: Input: nums = [4,5,6,7,0,1,2], target = 3 Output: -1 Example 3: Input: nums = [1], target = 0 Output: -1

Constraints:

1 <= nums.length <= 5000
-104 <= nums[i] <= 104
All values of nums are unique.
nums is an ascending array that is possibly rotated.
-104 <= target <= 104

Approach & Solution Steps

• Brute force — O(n) — scan.
• Better — find pivot then vanilla BS — two passes still O(log n) if careful.
• Optimal — O(log n) time / O(1) space — single binary search: decide left sorted vs right sorted half and whether target is in-range.

Optimal JS Solution

function search(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;
		}
		if (nums[left] <= nums[mid]) {
			if (nums[left] <= target && target < nums[mid]) {
				right = mid - 1;
			} else {
				left = mid + 1;
			}
		} else {
			if (nums[mid] < target && target <= nums[right]) {
				left = mid + 1;
			} else {
				right = mid - 1;
			}
		}
	}
	return -1;
}

Edge Cases & Pitfalls

• Always consider empty or null inputs.
• Watch out for off-by-one index errors.