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 rotated4
times. -
[0,1,2,4,5,6,7]
if it was rotated7
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:
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.
Constraints:
-
n == nums.length
-
1 <= n <= 5000
-
-5000 <= nums[i] <= 5000
- All the integers of
nums
are unique. -
nums
is sorted and rotated between1
andn
times.
SOLUTION:
class Solution:
def findMin(self, nums: List[int]) -> int:
n = len(nums)
beg = 0
end = n
while beg <= end:
mid = (beg + end) // 2
if mid < n - 1 and nums[mid] > nums[mid + 1]:
return nums[mid + 1]
if beg >= end - 1:
return nums[0]
elif nums[mid] > nums[0]:
beg = mid
else:
end = mid
Top comments (0)