153. Find Minimum in Rotated Sorted Array

Medium

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.

Solution

class Solution:
    def findMin(self, nums: List[int]) -> int:
        return self.helper(nums, 0, len(nums) - 1)

    def helper(self, nums: List[int], low: int, high: int) -> int:
        if low > high:
            return nums[0]

        if low == high:
            return nums[low]

        mid = (high + low) // 2

        if nums[mid] > nums[mid + 1]:
            return nums[mid + 1]

        if nums[mid - 1] > nums[mid]:
            return nums[mid]

        if nums[mid] < nums[high]:
            return self.helper(nums, low, mid - 1)

        return self.helper(nums, mid + 1, high)