153. Find Minimum in Rotated Sorted Array (Medium)

Description:

Suppose an array sorted in ascending order is rotated at some pivot unknown to you beforehand.

(i.e., [0,1,2,4,5,6,7] might become [4,5,6,7,0,1,2]).

Find the minimum element.

You may assume no duplicate exists in the array.

Example 1:

Input: [3,4,5,1,2]
Output: 1

Example 2:

Input: [4,5,6,7,0,1,2]
Output: 0

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Solutions:

NOTE: 二分查找

• length == 1 or 2 的时候单独考虑比较稳妥，不容易出错。
• leftmost < rightmost => sorted => return leftmost
• mid+1 as one of the next boundaries

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

Runtime: 48 ms, faster than 73.14% of Python3 online submissions for Find Minimum in Rotated Sorted Array.
Memory Usage: 14 MB, less than 5.40% of Python3 online submissions for Find Minimum in Rotated Sorted Array.

Recursive solution: inspired by https://www.youtube.com/watch?v=P4r7mF1Jd50

https://youtu.be/P4r7mF1Jd50?t=379

No need to use mid+1 as boundary, since both sides are considered, and at least one side can return the result in O(1) time (at least one side is sorted).

class Solution:
    def findMin(self, nums: List[int]) -> int:
        
        def search(l,r):
            if l == r-1 or nums[l] < nums[r-1]:
                return nums[l]
            if l == r-2:
                return min(nums[l],nums[l+1])
            
            mid = (l+r)//2
            return min(search(l,mid),search(mid,r))
    
        return search(0,len(nums))

Runtime: 52 ms, faster than 37.40% of Python3 online submissions for Find Minimum in Rotated Sorted Array.
Memory Usage: 14 MB, less than 5.40% of Python3 online submissions for Find Minimum in Rotated Sorted Array.

class Solution:
    def findMin(self, nums: List[int]) -> int:
        
        def search(l,r):
            if l == r-1 or nums[l] < nums[r-1]:
                return nums[l]
            
            mid = (l+r)//2
            return min(search(l,mid),search(mid,r))
    
        return search(0,len(nums))

Runtime: 48 ms, faster than 73.14% of Python3 online submissions for Find Minimum in Rotated Sorted Array.
Memory Usage: 14 MB, less than 5.40% of Python3 online submissions for Find Minimum in Rotated Sorted Array.