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Given an array 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.

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

Note: You are not suppose to use the library's sort function for this problem.

Example:

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

Follow up:

  • A rather straight forward solution is a two-pass algorithm using counting sort.
    First, iterate the array counting number of 0's, 1's, and 2's, then overwrite array with total number of 0's, then 1's and followed by 2's.
  • Could you come up with a one-pass algorithm using only constant space?

1 Answer

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by Goeduhub's Expert (2.3k points)
 
Best answer

We are classifying the array into four groups: red, white, unclassified, and blue. Initially we group all elements into unclassified. We iterate from the beginning as long as the white pointer is less than the blue pointer.

If the white pointer is red (nums[white] == 0), we swap with the red pointer and move both white and red pointer forward. If the pointer is white (nums[white] == 1), the element is already in correct place, so we don't have to swap, just move the white pointer forward. If the white pointer is blue, we swap with the latest unclassified element.

class Solution:

    def sortColors(self, nums: List[int]) -> None:

        """

        Do not return anything, modify nums in-place instead.

        """

        red, white, blue = 0, 0, len(nums)-1

    

        while white <= blue:

            if nums[white] == 0:

                nums[red], nums[white] = nums[white], nums[red]

                white += 1

                red += 1

            elif nums[white] == 1:

                white += 1

            else:

                nums[white], nums[blue] = nums[blue], nums[white]

                blue -= 1

eg : ans = Solution()

ans.sortColors([2,0,2,1,1,0])

Output -

[0,0,1,1,2,2]

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