Binary Search - Search in Rotated Sorted Array

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

Search in Rotated Sorted Array - 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

 

def bs_rotated(nums, target):
    def bs(lst, start, end):
        if start > end:
            return -1

        mid = (start + end) // 2
        if lst[mid] < target:
            return bs(lst, mid + 1, end)
        elif lst[mid] > target:
            return bs(lst, start, mid - 1)
        else:
            return mid

    if not nums:
        return -1

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

    added = nums + nums[:left]

    result = bs(added, left, len(added) - 1)
    return result if result == -1 else result % len(nums)

assert bs_rotated(nums=[4,5,6,7,0,1,2], target=0) == 4