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 1:

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

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

• 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?

## Solution

from typing import List

class Solution:
def sortColors(self, nums: List[int]) -> None:
redPtr = 0
bluePtr = len(nums) - 1
i = 0
while i <= bluePtr:
if nums[i] == 0:
nums[i], nums[redPtr] = nums[redPtr], nums[i]
redPtr += 1
i += 1
elif nums[i] == 2:
nums[i], nums[bluePtr] = nums[bluePtr], nums[i]
bluePtr -= 1
else:
i += 1


## Test Cases

test = Solution()
arr = [2,0,2,1,1,0]
test.sortColors(arr)
assert arr == [0,0,1,1,2,2],arr

arr = [2,0,2,1,1,2]
test.sortColors(arr)
assert arr == [0,1,1,2,2,2]

arr = []
test.sortColors(arr)
assert arr == []

arr = [0]
test.sortColors(arr)
assert arr == [0]

arr = [1]
test.sortColors(arr)
assert arr == [1]

arr = [2]
test.sortColors(arr)
assert arr == [2]

arr = [2,1,0]
test.sortColors(arr)
assert arr == [0,1,2]

arr = [2,0,2,1,1,0,2,1,1,0,2,1,1,0,2,1,1,0,2,1,0,2,1,1,1,0,2,1,1,0,2,1,1,0,2,1,0,2,1,0,2,1,0,2,0,2,1,0,2,1,1,1,1,1,1,1,1,0]
test.sortColors(arr)
assert arr == [0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2]
print('All Passed!')


## Big O Analysis

Space Complexity: O(1)

Time Complexity: O(N)