153 Find Minimum in Rotated Sorted Array

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.

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

        def helper(l: int, r: int) -> int:
            if nums[r] > nums[l]:
                return nums[l]

            mid = l + (r - l) // 2
            if nums[mid] > nums[mid + 1]:
                return nums[mid + 1]
            
            if nums[mid-1] > nums[mid]:
                return nums[mid]
            
            if nums[mid] > nums[l]:
                return helper(mid+1, r)
            else:
                return helper(l, mid-1)

        return helper(0, len(nums) - 1)