33. Search in Rotated Sorted Array

Medium LeetCode

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
• -10^4 <= nums[i] <= 10^4
• All values of nums are unique.
• nums is an ascending array that is possibly rotated.
• -10^4 <= target <= 10^4

How to solve the problem

• The key point is to find out which part is sorted and target is in which part of the nums array.

Code

• Binary Search

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

        while left <= right:
            mid = (left + right) // 2
            if nums[mid] == target:
                return mid

            if nums[left] <= nums[mid]: # which means left part is sorted
                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

Complexity

Time complexity: O(log n)

Space complexity: O(1)