Search in Rotated Sorted Array

Problem Link

Approach

Binary search while identifying which half remains sorted.

At each mid:

• If nums[mid] == target, return mid.
• If the left half [left, mid] is sorted, check whether target lies in that range. If yes, move right to mid minus 1. Otherwise search the right half.
• Otherwise the right half is sorted. Apply the same logic.

At least one half is always sorted in a rotated array with distinct values.

Linear scan is O(n). Binary search stays O(log n).

Time Complexity: O(log n)
Space Complexity: O(1)

Code

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 nums[mid] == target:
                return mid

            if nums[left] <= nums[mid]:
                if nums[left] <= target < nums[mid]:
                    right = mid - 1
                else:
                    left = mid + 1
            else:
                if nums[mid] < target <= nums[right]:
                    left = mid + 1
                else:
                    right = mid - 1

        return -1