Leetcode #153
pretty straightforward.
Problem
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.
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 between 1 and n times.
My Solution:
class Solution:
def findMin(self, nums: List[int]) -> int:
"""
- binary search for the min
- if initial nums[l] < nums[r]
- sorted
- return nums[0]
- while l < r:
- if nums[m] > nums[r]
- explore right
- l = m+1
- elif nums[m] <= nums[r]
- explore left
- r = m-1
"""
l,r = 0, len(nums)-1
if nums[l] < nums[r]:
return nums[l]
while l < r:
m = l + (r-l)//2
if nums[m] > nums[r]:
l = m+1
else:
r = m
return nums[l]