🚀Day 1. Sort 0s, 1s, and 2s 🧠

The problem can be found at the following link: Problem Link

💡 Problem Description:

You are given an array arr[] containing only 0s, 1s, and 2s. The task is to sort the array in ascending order. The problem must be solved in-place, and the space complexity should be constant.

🔍 Example Walkthrough:

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

Explanation: The array is sorted into ascending order.

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

Explanation: The array is sorted into ascending order.

Constraints:

• 1 <= arr.size() <= 10^6

• 0 <= arr[i] <= 2

🎯 My Approach:

1. Dutch National Flag Algorithm:

   • This problem can be solved using a two-pointer approach, also known as the Dutch National Flag Algorithm.

   • It divides the array into three sections:

     • All 0s will be placed at the beginning.

     • All 2s will be placed at the end.

     • All 1s will remain in the middle.

   • By iterating through the array and performing swaps, we can sort the array in a single pass.

2. Steps:

   • Initialize three pointers: low, mid, and high.

   • Use low to track the position of 0s, high to track the position of 2s, and mid to traverse the array.

   • Swap elements as necessary to place 0s, 1s, and 2s in their respective positions.

   • Continue until mid crosses high.

🕒 Time and Auxiliary Space Complexity

• Expected Time Complexity: O(n), as we iterate through the array exactly once, performing constant-time operations during each step.

• Expected Auxiliary Space Complexity: O(1), as we use only three pointers (low, mid, high) and no extra data structures.

📝 Solution Code

Code (C)

void sort012(int arr[], int n) {
    int low = 0, mid = 0, high = n - 1;

    while (mid <= high) {
        switch (arr[mid]) {
            case 0:
                {
                    int temp = arr[low];
                    arr[low] = arr[mid];
                    arr[mid] = temp;
                }
                low++;
                mid++;
                break;

            case 1:
                mid++;
                break;

            case 2:
                {
                    int temp = arr[high];
                    arr[high] = arr[mid];
                    arr[mid] = temp;
                }
                high--;
                break;
        }
    }
}

Code (Cpp)

class Solution {
public:
    void sort012(vector<int>& arr) {
        int low = 0, mid = 0, high = arr.size() - 1;

        while (mid <= high) {
            switch (arr[mid]) {
                case 0:
                    swap(arr[mid++], arr[low++]);
                    break;
                case 1:
                    mid++;
                    break;
                case 2:
                    swap(arr[mid], arr[high--]);
                    break;
            }
        }
    }
};

Code (Java)

class Solution {
    public void sort012(int[] arr) {
        int low = 0, mid = 0, high = arr.length - 1;

        while (mid <= high) {
            switch (arr[mid]) {
                case 0:
                    int temp0 = arr[low];
                    arr[low] = arr[mid];
                    arr[mid] = temp0;
                    low++;
                    mid++;
                    break;

                case 1:
                    mid++;
                    break;

                case 2:
                    int temp2 = arr[high];
                    arr[high] = arr[mid];
                    arr[mid] = temp2;
                    high--;
                    break;
            }
        }
    }
}

Code (Python)

class Solution:
    def sort012(self, arr):
        low, mid, high = 0, 0, len(arr) - 1

        while mid <= high:
            if arr[mid] == 0:
                arr[low], arr[mid] = arr[mid], arr[low]
                low += 1
                mid += 1
            elif arr[mid] == 1:
                mid += 1
            else:
                arr[mid], arr[high] = arr[high], arr[mid]
                high -= 1

🎯 Contribution and Support:

For discussions, questions, or doubts related to this solution, feel free to connect on LinkedIn: Any Questions. Let’s make this learning journey more collaborative!

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