# 153. Find minimum in rotated sorted array

Find Minimum in Rotated Sorted Array - LeetCode

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, a, a, ..., a[n-1]] 1 time results in the array [a[n-1], a, a, a, ..., 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.

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

return nums[left]

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

return nums[left]
``````