Open In App

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

Last Updated : 22 Nov, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

View all POTD Solutions

Welcome to the daily solutions of our PROBLEM OF THE DAY (POTD). We will discuss the entire problem step-by-step and work towards developing an optimized solution. This will not only help you brush up on your concepts of Matrix but will also help you build up problem-solving skills.

8th-november

We recommend you to try this problem on our GeeksforGeeks Practice portal first, and maintain your streak to earn Geeksbits and other exciting prizes, before moving towards the solution.

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:

C++




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;
    }
};


Java




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;
    }
}


Python3




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)



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads