LeetCode 33. Search in Rotated Sorted Array

Description

https://leetcode.com/problems/search-in-rotated-sorted-array/

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

You are given a target value to search. If found in the array return its index, otherwise return -1.

You may assume no duplicate exists in the array.

Your algorithm’s runtime complexity must be in the order of O(log n).

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

Explanation

find the rotation point and then do binary search

Python Solution

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        result = -1
        if nums == None or len(nums) == 0:
            return -1
        
        min_index = self.find_min_index(nums)
        
        if nums[min_index] <= target <= nums[-1]:
            return self.binary_search(nums, target, min_index, len(nums) - 1)
        
        return self.binary_search(nums, target, 0, min_index)

        
    def find_min_index(self, nums):
        start = 0
        end = len(nums) - 1
                
        while start + 1 < end:
            mid = start + (end - start) // 2

            if nums[mid] < nums[end]:
                end = mid
            else:
                start = mid
                
        if nums[start] < nums[end]:
            return start
        
        return end
    
    def binary_search(self, nums, target, start, end):
        while start + 1 < end:
            mid = start + (end - start) // 2

            if nums[mid] == target:
                return mid            
            elif nums[mid] < target:
                start = mid
            else:
                end = mid
                
        if nums[start] == target:
            return start
        elif nums[end] == target:
            return end
        
        return -1

• Time complexity: O(N).
• Space complexity: O(1).