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0154. Find Minimum In Rotated Sorted Array Ii

154. Find Minimum in Rotated Sorted Array II

Hard


Suppose an array of length n sorted in ascending order is rotated between 1 and n times. For example, the array nums = [0,1,4,4,5,6,7] might become:

  • [4,5,6,7,0,1,4] if it was rotated 4 times.
  • [0,1,4,4,5,6,7] 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 that may contain duplicates, return the minimum element of this array.

You must decrease the overall operation steps as much as possible.

 

Example 1:

Input: nums = [1,3,5]
Output: 1

Example 2:

Input: nums = [2,2,2,0,1]
Output: 0

 

Constraints:

  • n == nums.length
  • 1 <= n <= 5000
  • -5000 <= nums[i] <= 5000
  • nums is sorted and rotated between 1 and n times.

 

Follow up: This problem is similar to Find Minimum in Rotated Sorted Array, but nums may contain duplicates. Would this affect the runtime complexity? How and why?

 


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

        #return min(nums)

        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]

From python 3.7 dict guarantees that order will be kept as they inserted, and popitem will use LIFO order but we need FIFO type system. so we need OrderedDict which have popIten(last = T/F) for this req. One thing, next(iter(dict)) will return the first key of the dict

Python