[Leetcode/파이썬] 153. Find Minimum in Rotated Sorted Array

Find Minimum in Rotated Sorted Array

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Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,2,4,5,6,7] might become:

• [4,5,6,7,0,1,2] if it was rotated 4 times.
• [0,1,2,4,5,6,7] if it was rotated 7 times.

Notice that rotating an array [a[0], a[1], a[2], ..., a[n-1]] 1 time results in the array [a[n-1], a[0], a[1], a[2], ..., a[n-2]].

Given the sorted rotated array nums of unique elements, return the minimum element of this array.

You must write an algorithm that runs in O(log n) time.

 

Example 1:

Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

Input: nums = [4,5,6,7,0,1,2]
Output: 0
Explanation: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Example 3:

Input: nums = [11,13,15,17]
Output: 11
Explanation: The original array was [11,13,15,17] and it was rotated 4 times. 

 

Constraints:

• n == nums.length
• 1 <= n <= 5000
• -5000 <= nums[i] <= 5000
• All the integers of nums are unique.
• nums is sorted and rotated between 1 and n times.

 

class Solution:
    def findMin(self, nums: List[int]) -> int:
        # 반드시 시간 O(log n) 으로 풀이할 것 => 여기서 이분탐색 등 을 생각할 수 있었음
        # 답(최저값)을 몰라도 찾을 수 가 있나...?
        # 부등호 설정할 때 이상, 이하, 초과, 미만 설정이 어렵다.
        left = 0
        right = len(nums) - 1

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