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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 rotated4
times.[0,1,2,4,5,6,7]
if it was rotated7
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 between1
ynorte
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))