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:

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

Example 2:

 1 2 3 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:

 1 2 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.

See the problem

Solution

Python3 Solution:

 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 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

 1 2 3 4 5 6 7 8 9 10 11 [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]