index.md

33. Search in Rotated Sorted Array

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.

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

Solution

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left, right = 0, len(nums) - 1

        while left <= right:
            mid = (left + right) // 2

            # If the target is found, return its index
            if nums[mid] == target:
                return mid

            # Determine which part of the array to search in
            if nums[left] <= nums[mid]:  # Left part is sorted
                if nums[left] <= target < nums[mid]:  # Target is in the left part
                    right = mid - 1
                else:  # Target is in the right part
                    left = mid + 1
            else:  # Right part is sorted
                if nums[mid] < target <= nums[right]:  # Target is in the right part
                    left = mid + 1
                else:  # Target is in the left part
                    right = mid - 1

        # If the loop ends, the target is not in the array
        return -1

Thoughts

The time complexity of this algorithm is O(log n), as it uses a binary search. The space complexity is O(1), as it only uses a constant amount of extra space.