33. 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.

 

'''
[1] binary search)

idea::
1. first find minimum_idx
2. left=minimum_idx, right=(minimum_idx-1)%len(nums), do binary search

code::


-T/C: O(logn) # two-binary search. 
-S/C: O(1)
'''

class Solution:
    def search(self, nums: List[int], target: int) -> int:
        left=0
        right=len(nums)-1
        
        # find min_idx
        while left<right:
            mid=left+(right-left)//2
            
            if nums[mid]>nums[right]:
                left=mid+1
            else:
                right=mid
        
        # at this time. need to find what is smaller num
        if nums[left]>nums[right]:
            left=right
        
        # do binary search
        right=left+len(nums)-1  # imagine values higher than min_val are appended to nums
        
        while left<=right:
            mid=left+(right-left)//2
            mid_idx=mid % len(nums)
            if nums[mid_idx]==target:
                return mid_idx
            elif nums[mid_idx]>target:
                right=mid-1
            else:
                left=mid+1
                
        return -1