Leetcode 153
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.
Intuition
The simplest way is to go through the array - time complexity is $O(n)$.
Since it’s a sorted array (although rotated), we can consider the binary search to satisfy the $O(logn)$ complexity.
- If nums[start] < nums[end], it means that nums is the longest sorted array from start to end, then we can return nums[start]
- If nums[mid] <= nums[0], it means that the right half is sorted, and the minimum element is in the right half.
- Else the left half is sorted, the minimum element is in the left half.
Code Implementation
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class Solution:
def findMin(self, nums: List[int]) -> int:
start = 0
end = len(nums) - 1
while start < end:
if nums[start] < nums[end]:
return nums[start]
mid = (int)(start + (end - start) / 2)
if nums[mid] <= nums[0]:
start = mid + 1
else:
end = mid
return nums[end]
