153.findMinRotatedSortedArr.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.
#
#
# 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.

from typing import List


class Solution:
    def findMin(self, nums: List[int]) -> int:
        res = nums[0]  # arbitrary value
        l, r = 0, len(nums) - 1

        # running binary search
        while l <= r:
            # if the subarray is sorthes, the min is the left most element
            if nums[l] < nums[r]:
                res = min(res, nums[l])
                break

            # find the pivot
            m = (l + r) // 2
            res = min(res, nums[m])
            # check if nums[m] is in left portion, we then search to the right portion
            if nums[m] >= nums[l]:
                # move pivoit to the left
                l = m + 1  # search right
            else:
                r = m - 1  # search left

        return res