Traverse matrix in L shape

Given a N * M matrix. The task is to traverse the given matrix in L shape as shown in below image. 

Examples:

Input : n = 3, m = 3
a[][] = { { 1, 2, 3 },
          { 4, 5, 6 },
          { 7, 8, 9 } } 
Output : 1 4 7 8 9 2 5 6 3

Input : n = 3, m = 4
a[][] = { { 1, 2, 3 },
          { 4, 5, 6 },
          { 7, 8, 9 },
          { 10, 11, 12} } 
Output : 1 4 7 10 11 12 2 5 8 9 3 6 

Observe there will be m (number of columns) number of L shapes that need to be traverse. So we will traverse each L shape in two parts, first vertical (top to down) and then horizontal (left to right). 
To traverse in vertically, observe for each column j, 0 <= j <= m – 1, we need to traverse n – j elements vertically. So for each column j, traverse from a[0][j] to a[n-1-j][j]. 

Now, to traverse horizontally for each L shape, observe the corresponding row for each column j will be (n-1-j)th row and the first element will be (j+1)th element from the beginning of the row. So, for each L shape or for each column j, to traverse horizontally, traverse from a[n-1-j][j+1] to a[n-1-j][m-1].

Below is the implementation of this approach:  

C++

// C++ program to traverse a m x n matrix in L shape.
#include <iostream>
using namespace std;

#define MAX 100

// Printing matrix in L shape
void traverseLshape(int a[][MAX], int n, int m)
{
    // for each column or each L shape
    for (int j = 0; j < m; j++) {

        // traversing vertically
        for (int i = 0; i <= n - j - 1; i++)
            cout << a[i][j] << " ";

        // traverse horizontally
        for (int k = j + 1; k < m; k++)
            cout << a[n - 1 - j][k] << " ";
    }
}

// Driver Program
int main()
{
    int n = 4;
    int m = 3;
    int a[][MAX] = { { 1, 2, 3 },
                     { 4, 5, 6 },
                     { 7, 8, 9 },
                     { 10, 11, 12 } };
    traverseLshape(a, n, m);
    return 0;
}

Java

// Java Program to traverse a m x n matrix in L shape.
public class GFG{

    static void traverseLshape(int a[][], int n, int m) {
        // for each column or each L shape
        for (int j = 0; j < m; j++) {

            // traversing vertically
            for (int i = 0; i <= n - j - 1; i++)
                System.out.print(a[i][j] + " ");

            // traverse horizontally
            for (int k = j + 1; k < m; k++)
                System.out.print(a[n - 1 - j][k] + " ");
        }
    }

    // Driver Code
    public static void main(String args[]) {
        int n = 4;
        int m = 3;
        int a[][] = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 },
                        { 10, 11, 12 } };
        traverseLshape(a, n, m);
    }
}

Python3

# Python3 program to traverse a
# m x n matrix in L shape.

# Printing matrix in L shape
def traverseLshape(a, n, m):

    # for each column or each L shape
    for j in range(0, m):

        # traversing vertically
        for i in range(0, n - j):
            print(a[i][j], end = " ");

        # traverse horizontally
        for k in range(j + 1, m):
            print(a[n - 1 - j][k], end = " ");

# Driven Code
if __name__ == '__main__':
    n = 4;
    m = 3;
    a = [[1, 2, 3],
         [4, 5, 6],
         [7, 8, 9],
         [10, 11, 12]];
    traverseLshape(a, n, m);

# This code is contributed by PrinciRaj1992

C#

// C# Program to traverse a m x n matrix in L shape.

using System;

public class GFG{

    static void traverseLshape(int[,] a, int n, int m) {
        // for each column or each L shape
        for (int j = 0; j < m; j++) {

            // traversing vertically
            for (int i = 0; i <= n - j - 1; i++)
                Console.Write(a[i,j] + " ");

            // traverse horizontally
            for (int k = j + 1; k < m; k++)
                Console.Write(a[n - 1 - j,k] + " ");
        }
    }

    // Driver Code
    public static void Main() {
        int n = 4;
        int m = 3;
        int[,] a = { { 1, 2, 3 },
                        { 4, 5, 6 },
                        { 7, 8, 9 },
                        { 10, 11, 12 } };
        traverseLshape(a, n, m);
    }
}

Javascript

<script>

// Javascript program to traverse a m x n matrix in L shape.

var MAX = 100;

// Printing matrix in L shape
function traverseLshape(a, n, m)
{
    // for each column or each L shape
    for (var j = 0; j < m; j++) {

        // traversing vertically
        for (var i = 0; i <= n - j - 1; i++)
            document.write( a[i][j] + " ");

        // traverse horizontally
        for (var k = j + 1; k < m; k++)
            document.write( a[n - 1 - j][k] + " ");
    }
}

// Driver Program
var n = 4;
var m = 3;
var a = [ [ 1, 2, 3 ],
                 [ 4, 5, 6 ],
                 [ 7, 8, 9 ],
                 [ 10, 11, 12 ] ];
traverseLshape(a, n, m);



</script>

Output

1 4 7 10 11 12 2 5 8 9 3 6 

Complexity Analysis:

• Time Complexity: O(n * m)
• Auxiliary Space: O(1)