# 000033-Search-in-Rotated-Sorted-Array

### Problem

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

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

binary search

``````class Solution:
def search(self, nums: List[int], target: int) -> int:
if not nums:
return -1

left, right = 0, len(nums) - 1
while left < right:
mid = (left+right)//2
if nums[mid] == target:
return mid
if nums[mid] >= nums[left]:
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 left if nums[left] == target else -1``````

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