Given an array nums with n objects colored red, white, or blue, sort them in-place so that objects of the same color are adjacent, with the colors in the order red, white, and blue.

We will use the integers 0, 1, and 2 to represent the color red, white, and blue, respectively.

You must solve this problem without using the library's sort function.

 

Example 1:

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

Example 2:

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

 

Constraints:

  • n == nums.length
  • 1 <= n <= 300
  • nums[i] is either 0, 1, or 2.

 

Follow up: Could you come up with a one-pass algorithm using only constant extra space?





 class Solution:
    def sortColors(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
#         for i in range(len(nums)):
#             min_index = i

#             for j in range(i+1, len(nums)):
#                 if nums[min_index] > nums[j]:
#                     min_index = j

#             nums[min_index], nums[i] =   nums[i] , nums[min_index]


        # using counting sort
        M = 2
        counter = [0]*(M+1)

        for item in nums:
            counter[item] += 1

        starting_index = 0
        for i, count in enumerate(counter):
            counter[i] = starting_index
            starting_index += count

        sorted_list = [0]*len(nums)
        for item in nums:
            sorted_list[counter[item]] = item
            counter[item]+=1

        for i in range(len(nums)):
            nums[i] = sorted_list[i]

Random Note


Amortized time is the way to express the time complexity when an algorithm has the very bad time complexity only once in a while besides the time complexity that happens most of time. Good example would be an ArrayList which is a data structure that contains an array and can be extended.