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 ` `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

 `    `

Output

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

Complexity Analysis:

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

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