Description
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- 4times.
- [0,1,2,4,5,6,7]if it was rotated- 7times.
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 inΒ O(log n) time.
Β
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.
Β
Constraints:
- n == nums.length
- 1 <= n <= 5000
- -5000 <= nums[i] <= 5000
- All the integers of numsare unique.
- numsis sorted and rotated between- 1and- ntimes.
Solution
Python3
class Solution:
    def findMin(self, nums: List[int]) -> int:
        N = len(nums)
        left, right = 0, N - 1
 
        while left < right:
            if nums[left] < nums[right]:
                return nums[left]
                
            mid = left + (right - left) // 2
 
            if nums[left] <= nums[mid]:
                left = mid + 1
            else:
                right = mid
        
        return nums[left]