Given an n x n matrix .In the given matrix, you have to print the elements of the matrix in the snake 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
We traverse all rows. For every row, we check if it is even or odd. If even, we print from left to right else print from right to left.
C++
// C++ program to print matrix in snake order #include <iostream> #define M 4 #define N 4 using namespace std; void print( int mat[M][N]) { // 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++) cout << mat[i][j] << " " ; // If current row is odd, print from // right to left } else { for ( int j = N - 1; j >= 0; j--) cout << mat[i][j] << " " ; } } } // Driver code int main() { int mat[][] = { { 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 } }; print(mat); return 0; } |
Java
// Java program to print matrix in snake order import java.util.*; class GFG { static void print( int [][] mat) { // Traverse through all rows for ( int i = 0 ; i < mat.length; i++) { // If current row is even, print from // left to right if (i % 2 == 0 ) { for ( int j = 0 ; j < mat[ 0 ].length; j++) System.out.print(mat[i][j] + " " ); // If current row is odd, print from // right to left } else { for ( int j = mat[ 0 ].length - 1 ; j >= 0 ; j--) System.out.print(mat[i][j] + " " ); } } } // Driver code public static void main(String[] args) { int mat[][] = new int [][] { { 10 , 20 , 30 , 40 }, { 15 , 25 , 35 , 45 }, { 27 , 29 , 37 , 48 }, { 32 , 33 , 39 , 50 } }; print(mat); } } /* This code is contributed by Mr. Somesh Awasthi */ |
Python 3
# Python 3 program to print # matrix in snake order M = 4 N = 4 def printf(mat): global M, N # Traverse through all rows for i in range (M): # If current row is # even, print from # left to right if i % 2 = = 0 : for j in range (N): print ( str (mat[i][j]), end = " " ) # If current row is # odd, print from # right to left else : for j in range (N - 1 , - 1 , - 1 ): print ( str (mat[i][j]), end = " " ) # Driver code mat = [[ 10 , 20 , 30 , 40 ], [ 15 , 25 , 35 , 45 ], [ 27 , 29 , 37 , 48 ], [ 32 , 33 , 39 , 50 ]] printf(mat) # This code is contributed # by ChitraNayal |
C#
// C# program to print // matrix in snake order using System; class GFG { static void print( int [,]mat) { // Traverse through all rows for ( int i = 0; i < mat.GetLength(0); i++) { // If current row is // even, print from // left to right if (i % 2 == 0) { for ( int j = 0; j < mat.GetLength(1); j++) Console.Write(mat[i, j] + " " ); // If current row is // odd, print from // right to left } else { for ( int j = mat.GetLength(1) - 1; j >= 0; j--) Console.Write(mat[i, j] + " " ); } } } // Driver code public static void Main() { int [,]mat = {{ 10, 20, 30, 40 }, { 15, 25, 35, 45 }, { 27, 29, 37, 48 }, { 32, 33, 39, 50 }}; print(mat); } } // This code is contributed // by ChitraNayal |
PHP
<?php // PHP program to print // matrix in snake order $M = 4; $N =4; function printLN( $mat ) { global $M ; global $N ; // Traverse through all rows for ( $i = 0; $i < $M ; $i ++) { // If current row is even, // print from left to right if ( $i % 2 == 0) { for ( $j = 0; $j < $N ; $j ++) echo $mat [ $i ][ $j ], " " ; // If current row is odd, // print from right to left } else { for ( $j = $N - 1; $j >= 0; $j --) echo $mat [ $i ][ $j ] , " " ; } } } // Driver code $mat = array ( array (10, 20, 30, 40), array (15, 25, 35, 45), array (27, 29, 37, 48), array (32, 33, 39, 50)); printLN( $mat ); // This code is contributed by ajit ?> |
Output :
10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
leave a comment
0 Comments