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 shapevoid 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 Programint 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 shapedef 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 Codeif __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 shapefunction 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 Programvar n = 4;var m = 3;var a = [ [ 1, 2, 3 ], [ 4, 5, 6 ], [ 7, 8, 9 ], [ 10, 11, 12 ] ];traverseLshape(a, n, m);</script> |
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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