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++

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// 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] << " ";
    }
}
  
// Driven 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;
}

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Java

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// 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); 
    }
}

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Python3

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# 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

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C#

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// 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);  
    

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Output:

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


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