leetcode/work/Medium/Array/153.find-minimum-in-rotated-sorted-array.py

"""
153. Find Minimum in Rotated Sorted Array
Difficulty: Medium
https://leetcode.com/problems/find-minimum-in-rotated-sorted-array/

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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 findMin(self, nums: List[int]) -> int:
        l, r = 0, len(nums) - 1
        lowest_index = -1

        while l <= r:
            m = (l + r) // 2
            if nums[m] <= nums[-1]:
                lowest_index = m
                r = m - 1
            else:
                l = m + 1

        return nums[lowest_index]