# 75. Sort Colors

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?

# Solution

Approach 1: Two passes - counting sort.

Approach 2: One pass - three pointers.

# Code (Python)

Approach 1:

from collections import Counter

    def sortColors(self, nums):
        counts = Counter(nums)
        i = 0
        for num in (0, 1, 2):
            for j in range(counts[num]):
                nums[i] = num
                i += 1

Approach 2:

    def sortColors(self, nums):
        pointer_0, pointer_2, current = 0, len(nums) - 1, 0
        while current <= pointer_2:
            if nums[current] == 0:
                nums[current], nums[pointer_0] = nums[pointer_0], 0
                pointer_0 += 1
                
                # Either of the following line works. Want to make sure current >= pointer_0 always,
                # and if current > pointer_0, the array looks like [0, 0, 0, 1, 1, 0, ...], so after the swap current always points to a 1
                #                                                 (pointer_0)|     |(current)

                #current += 1 # OR:
                current = max(current, pointer_0)
            elif nums[current] == 2:
                nums[current], nums[pointer_2] = nums[pointer_2], 2
                pointer_2 -= 1
            else:
                current += 1

# Code (C++)

Approach 1:

// Two passes - counting sort.
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int colorCount[3] = {0};
        for (int i = 0; i < nums.size(); ++i)
        {
            colorCount[nums[i]]++;
        }
        int i = 0;
        for (int j = 0; j < 3; ++j)
        {
            for (int k = 0; k < colorCount[j]; ++k)
            {
                nums[i++] = j;
            }
        }
    }
};

Approach 2:

// One pass - three pointers.
class Solution {
public:
    void sortColors(vector<int>& nums) {
        int p0 = 0;
        int p1 = 0;
        int p2 = nums.size() - 1;
        while (p0 < nums.size() && p1 < nums.size() && p2 >= 0)
        {
            if (nums[p0] == 0)
                p0++;
            else if (nums[p2] == 2)
                p2--;
            else if (p1 > p0 && nums[p1] == 0)
                std::swap(nums[p1], nums[p0]);
            else if (p1 < p2 && nums[p1] == 2)
                std::swap(nums[p1], nums[p2]);
            else
                p1++;
        }
    }
};