[Leetcode] 154. Find Minimum in Rotated Sorted Array II

Problem:

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

[4,5,6,7,0,1,4] if it was rotated 4 times.
[0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

 

Example 1:

Input: nums = [1,3,5]
Output: 1
Example 2:

Input: nums = [2,2,2,0,1]
Output: 0
 

Constraints:

n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
nums is sorted and rotated between 1 and n times.

 

Solution:

class Solution:
    def findMin(self, nums: List[int]) -> int:
        left,right = 0 , len(nums) -1 
        mid = 0 
        while left < right:
            mid = (left + right) // 2 
            if nums[mid] == nums[right]:
                right -= 1 
            elif nums[mid] < nums[right]:
                right = mid 
            else:
                left = mid + 1 
        return nums[left]

위 Solution은 다른 사람의 Solution을 참고하여 작성했다. duplicates가 있기 때문에 Binary Search를 어떻게 써야 할지 고민을 많이 했다. 내가 생각하지 못한 부분은 nums [mid]와 nums [right]가 같을 때, right -= 1을 해주는 부분이다. 이때 내가 깨달은 것은 right에 1을 빼주면 mid값도 순차적으로 줄어들어서 모든 경우의 수를 경험한다는 것이다.