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
1 <= m, n <= 100
nums[i] is either 0, 1, or 2

Approach 1: Counting

Count number of 0's 1's and 2's and add those numbers in array

class Solution {
    public void sortColors(int[] nums) {
        int q[] = { 0, 0, 0 };
        for (int i : nums) {
            if (i == 0) {
                q[0]++;
            } else if (i == 1) {
                q[1]++;
            } else {
                q[2]++;
            }
        }
        for (int i = 0; i < q[0]; i++) {
            nums[i] = 0;
        }
        for (int i = q[0]; i < q[0] + q[1]; i++) {
            nums[i] = 1;
        }
        for (int i = q[1] + q[0]; i < q[0] + q[1] + q[2]; i++) {
            nums[i] = 2;
        }
    }
}

Approach 2: Bubble Sort

We want all integers together. One way of doing this is sorting the array so that we will get all $0$ initially, then all $1$ and $2$ . We can use bubble sort for sorting.

Time Complexity: $O(n ^ 2)$ , where $n$ is the number of elements in array

Space complexity: $O(1)$

class Solution {
    void sortColors(vector<int>& nums) {
        int n = nums.size();
        bool check = true;
        for (int i = 0; i < n - 1 && check; i++) {
            check = false;
            for (int j = 0; j < n - i - 1; j++) {
                if (nums[j] > nums[j + 1]) {
                    swap(nums[j], nums[j + 1]);
                    check = true;
                }
            }
        }
    }
};

/**
 * @param {number[]} nums
 * @return {void} Do not return anything, modify nums in-place instead.
 */
var sortColors = function (nums) {
  let swapped = false;
  for (i = 0; i < nums.length - 1; i++) {
    for (j = 0; j < nums.length - i - 1; j++) {
      if (nums[j] > nums[j + 1]) {
        swapped = true;
        let tmp = nums[j];
        nums[j] = nums[j + 1];
        nums[j + 1] = tmp;
      }
    }
    if (swapped === false) {
      return;
    }
  }
};

class Solution:
    def sortColors(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        swapped = False
        for i in range(len(nums) - 1):
            for j in range(len(nums) - i - 1):
                if nums[j] > nums[j + 1]:
                    swapped = True
                    nums[j], nums[j + 1] = nums[j + 1], nums[j]
            if not swapped:
                return

0075 - Sort Colors (Medium)

Problem Link

Problem Statement

Approach 1: Counting

Approach 2: Bubble Sort

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