# 153. Find Minimum in Rotated Sorted Array """ 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' mine become a different array. return the sorted array unique elements minimum times. algorithm must be o log n time. """ class Solution(object): def findMin(self, nums): """ :type nums: List[int] :rtype: int """ if len(nums) == 1 or nums[0] < nums[-1]: return nums[0] left, right = 0, len(nums) - 1 while left <= right: mid = left + (right - left) // 2 if mid > 0 and nums[mid - 1] > nums[mid]: return nums[mid] if nums[mid] > nums[right]: left = mid + 1 else: right = mid - 1 if __name__ == "__main__": obj = Solution() print(obj.findMin([3, 4, 5, 1, 2])) print(obj.findMin([4, 5, 6, 7, 0, 1, 2])) print(obj.findMin([11, 13, 15, 17]))