[LeetCode] 0153. Find Minimum in Rotated Sorted Array

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Medio

Suppose an array of length norte sorted in ascending order is rotated between 1 y norte 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 una matriz [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 números of unique elements, return the minimum element of this array.

Debes escribir un algoritmo que se ejecute en O(log n) time.

Ejemplo 1:

Aporte: nums = [3,4,5,1,2]
Producción: 1
Explicación: The original array was [1,2,3,4,5] rotated 3 times.

Ejemplo 2:

Aporte: nums = [4,5,6,7,0,1,2]
Producción: 0
Explicación: The original array was [0,1,2,4,5,6,7] and it was rotated 4 times.

Ejemplo 3:

Aporte: nums = [11,13,15,17]
Producción: 11
Explicación: The original array was [11,13,15,17] and it was rotated 4 times.

Restricciones:

n == nums.length
1 <= n <= 5000
-5000 <= nums[i] <= 5000
All the integers of números are unique.
números is sorted and rotated between 1 y norte times.

Pitón

				
					# time complexity: O(logn)
# space complexity: O(1)
from typing import List


class Solution:
    def findMin(self, nums: List[int]) -> int:
        if len(nums) == 1:
            return nums[0]
        left, right = 0, len(nums) - 1
        if nums[right] > nums[0]:
            return nums[0]
        while left <= right:
            mid = left + (right - left) // 2
            if nums[mid] > nums[mid + 1]:
                return nums[mid+1]
            if nums[mid] < nums[mid - 1]:
                return nums[mid]
            if nums[mid] > nums[-1]:
                left = mid + 1
            else:
                right = mid - 1
        return nums[mid]


nums = [11, 13, 15, 17]

print(Solution().findMin(nums))

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Pitón

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