Find Minimum in Rotated Sorted Array

Problem Statement

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:

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Input: nums = [3,4,5,1,2]
Output: 1
Explanation: The original array was [1,2,3,4,5] rotated 3 times.

Example 2:

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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:

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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.

See the problem

Solution

Python3 Solution:

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class Solution:
    def rotation_point(self, nums):
        n = len(nums)
        start = 0
        end = n - 1

        if n == 1:
            return 0

        while True:
            mid = (start + end) // 2

            if (
                nums[mid] <= nums[(mid + 1 + n) % n]
                and nums[mid] < nums[(mid - 1 + n) % n]
            ):
                return mid
            elif nums[mid] > nums[end]:
                start = mid + 1
            else:
                end = mid - 1

    def findMin(self, nums: List[int]) -> int:
        i = self.rotation_point(nums)
        return nums[i]

Test Cases

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[3,4,5,1,2]
[3,4,5,6, 7, 0,1, 2]
[1]
[1,2, 3, 4]
[5,1,2,3,4]
[2,3,4,5,1]
[1,2]
[2,1]
[1,2,3]
[3,1,2]
[2,3,1]