POTD Solutions | 8 Nov’ 23 | Print Matrix in snake Pattern

POTD 8 November: Print Matrix in snake Pattern

Given a matrix of size N x N. Print the elements of the matrix in the snake like pattern.

Examples : 

Input: mat[][] = { {10, 20, 30, 40},
                            {15, 25, 35, 45},
                           {27, 29, 37, 48},
                           {32, 33, 39, 50}};
Output: 10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32 

Input: mat[][] = { {1, 2, 3},
                             {4, 5, 6},
                             {7, 8, 9}};
Output: 1 2 3 6 5 4 7 8 9

Approach: Follow the steps below to solve the problem:

• Traverse all rows. 
• For every row, check if it is even or odd. 
• If even, we print from left to right 
• else print from right to left. 

Below is the implementation of above approach:

class Solution {
public:
    // Function to return list of integers visited in snake
    // pattern in matrix.
    vector<int> snakePattern(vector<vector<int> > matrix)
    {
        vector<int> ans;
        int N = matrix.size();
        int M = matrix[0].size();
        // Traverse through all rows
        for (int i = 0; i < M; i++) {

            // If current row is even, print from
            // left to right
            if (i % 2 == 0) {
                for (int j = 0; j < N; j++)
                    ans.push_back(matrix[i][j]);

                // If current row is odd, print from
                // right to left
            }
            else {
                for (int j = N - 1; j >= 0; j--)
                    ans.push_back(matrix[i][j]);
            }
        }
        return ans;
    }
};

class Solution {
    // Function to return list of integers visited in snake
    // pattern in matrix.
    static ArrayList<Integer> snakePattern(int matrix[][])
    {
        ArrayList<Integer> ans = new ArrayList<>();
        int N = matrix.length;
        int M = matrix[0].length;

        // Traverse through all rows
        for (int i = 0; i < M; i++) {
            // If the current row is even, print from left
            // to right
            if (i % 2 == 0) {
                for (int j = 0; j < N; j++) {
                    ans.add(matrix[i][j]);
                }
            }
            else { // If the current row is odd, print from
                   // right to left
                for (int j = N - 1; j >= 0; j--) {
                    ans.add(matrix[i][j]);
                }
            }
        }
        return ans;
    }
}

class Solution:
    
    #Function to return list of integers visited in snake pattern in matrix.
    def snakePattern(self, matrix):
        ans = []
        N = len(matrix)
        M = len(matrix[0])

        # Traverse through all rows
        for i in range(M):
            # If the current row is even, print from left to right
            if i % 2 == 0:
                for j in range(N):
                    ans.append(matrix[i][j])
            else:  # If the current row is odd, print from right to left
                for j in range(N - 1, -1, -1):
                    ans.append(matrix[i][j])

        return ans

Time Complexity: O(N x M), Traversing over all the elements of the matrix, therefore N X M elements are there.
Auxiliary Space: O(1)