find_minimum_in_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. 
 

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:
        if nums[0] <= nums[-1]:
            return nums[0]
      
        # Initialize the left and right pointers.
        left, right = 0, len(nums) - 1
      
        # Perform a binary search for the minimum element.
        while left < right:
            # Calculate the midpoint index
            mid = (left + right) // 2  # Using // for floor division in Python 3
          
            # If the element at the midpoint is greater than or equal 
            # to the first element, then the minimum is to the right.
            if nums[0] <= nums[mid]:
                left = mid + 1
            else:
                # Otherwise, the minimum is to the left, so we reduce the right bound.
                right = mid
      
        # After the loop, left will point to the smallest element.
        return nums[left]