[LeetCode] 0033. Search in Rotated Sorted Array

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Table of contents

Medium

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

Python

				
					from typing import List class Solution: def search(self, nums: List[int], target: int) -> int: # return nums.index(target) if target in nums else -1 n = len(nums) left, right = 0, n-1 while left <= right: mid = left + (right - left) // 2 if nums[mid] > nums[-1]: left = mid + 1 else: right = mid - 1 def binarySearch(left_boundary: int, right_boundary: int, target: int): left, right = left_boundary, right_boundary while left <= right: mid = left + (right - left) // 2 if nums[mid] == target: return mid elif nums[mid] > target: right = mid - 1 else: left = mid + 1 return -1 if (answer := binarySearch(0, left-1, target)) != -1: return answer return binarySearch( left, n-1, target)

Medium

Python

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