33. Search in Rotated Sorted Array

• categories: Code, Interview Question, leetcode, Medium
• source: https://leetcode.com/problems/search-in-rotated-sorted-array
• topics: Binary Search, Array

Description

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

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.

Idea

Use an iterative binary search. At each step:

1. Check if the target matches the middle element.
2. Determine if the left half is sorted. If sorted, check if the target lies in this range; if yes, search in the left half, otherwise, search in the right half.
3. If the left half is not sorted, the right half must be sorted. Repeat the same check for the target within the right half.
4. Continue narrowing the search range until the target is found or the range is empty.

This ensures O(log⁡n)O(\log n)O(logn) complexity by eliminating half the array at each ste

Code

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        if not nums:
            return -1
 
        left, right = 0, len(nums) - 1
        while left <= right:
            mid = (left + right) // 2
 
            if nums[mid] == target:
                return mid
 
            # Check if the left half is sorted
            if nums[left] <= nums[mid]:
                if nums[left] <= target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            # Right half must be sorted
            else:
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1
 
        return -1