Leetcode•Jun 24, 2025
Search in Rotated Sorted Array
Hazrat Ali
Leetcode
Prior to being passed to your function, nums is possibly rotated at an unknown pivot 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 rotated at pivot index 3 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
Solution
/**
* @param {number[]} nums
* @param {number} target
* @return {number}
*/
const search = (nums, target) => {
if (nums.length === 0) {
return -1;
}
let lo = 0;
let hi = nums.length - 1;
while (lo <= hi) {
const mid = lo + Math.floor((hi - lo) / 2);
if (nums[mid] === target) {
return mid;
}
if (nums[lo] <= nums[mid]) {
if (nums[lo] <= target && target < nums[mid]) {
hi = mid - 1;
} else {
lo = mid + 1;
}
} else {
if (nums[mid] < target && target <= nums[hi]) {
lo = mid + 1;
} else {
hi = mid - 1;
}
}
}
return -1;
};