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.

 class Solution:
    def intersection(self, nums, left, right):
        if nums[right] > nums[left]:
            return 0

        while left <= right:
            mid = left + (right-left)//2

            if nums[mid] > nums[mid+1]:
                 return mid + 1
            else:
                if nums[mid] < nums[left]:
                    right = mid - 1
                else:
                    left = mid + 1

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

Find Minimum In Rotated Sorted Array

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