0153. Find Minimum In Rotated Sorted Array

153. Find Minimum in Rotated Sorted Array

Medium

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.

 class Solution:
    def findMin(self, nums: List[int]) -> int:

#         if nums[0] < nums[-1]:
#             return nums[0]

#         left, right = 0, len(nums)-1

#         while left < right:
#             mid = (left+right)//2

#             if nums[mid] > nums[mid+1]:
#                 return nums[mid+1]

#             elif nums[mid] > nums[0]:
#                 left = mid+1
#             else:
#                 right = mid


#         return nums[left]

        left = 0
        right = len(nums)-1


        while left < right:

            mid = (left + right)//2

            if nums[right] < nums[mid]:
                left = mid+1
            elif nums[right] > nums[mid]:
                right = mid
            else:
                right -= 1



        return nums[left]

0153. Find Minimum In Rotated Sorted Array

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