find-minimum-in-rotated-sorted-array.py

"""
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.

"""
from typing import List
class Solution:
    def findMin(self, nums: List[int]) -> int:
        # The scenarios are:
        # Search right of mid:
        ## If mid > cur_min and r_val <= current_min

        # Search to the left of mid:
        ## if mid > cur_min and l_val <= current_min


        min = nums[0]

        l = 0
        r = len(nums) - 1

        while l <= r:
            mid = (l + r) // 2
            val_mid = nums[mid]

            if val_mid <= min:
                min = val_mid
            if nums[r] <= val_mid:
                l = mid + 1
            else:
                r = mid - 1

        return min

x = Solution()

print(x.findMin([11,13,15,17]))
print(x.findMin([4,5,6,7,0,1,2]))
print(x.findMin([6, 1, 2, 3, 4, 5]))