Print matrix in snake pattern
Last Updated :
19 Sep, 2022
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
Approach: Follow the steps below to solve the problem:
- Traverse all rows.
- For every row, check if it is even or odd.
- If even, we print from left to right
- else print from right to left.
Below is the implementation of above approach:
C++
#include <iostream>
#define M 4
#define N 4
using namespace std;
void print(int mat[M][N])
{
for (int i = 0; i < M; i++) {
if (i % 2 == 0) {
for (int j = 0; j < N; j++)
cout << mat[i][j] << " ";
}
else {
for (int j = N - 1; j >= 0; j--)
cout << mat[i][j] << " ";
}
}
}
int main()
{
int mat[M][N] = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
print(mat);
return 0;
}
|
Java
import java.util.*;
class GFG {
static void print(int[][] mat)
{
for (int i = 0; i < mat.length; i++) {
if (i % 2 == 0) {
for (int j = 0; j < mat[0].length; j++)
System.out.print(mat[i][j] + " ");
}
else {
for (int j = mat[0].length - 1; j >= 0; j--)
System.out.print(mat[i][j] + " ");
}
}
}
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);
}
}
|
Python3
M = 4
N = 4
def printf(mat):
global M, N
for i in range(M):
if i % 2 == 0:
for j in range(N):
print(str(mat[i][j]),
end=" ")
else:
for j in range(N - 1, -1, -1):
print(str(mat[i][j]),
end=" ")
mat = [[10, 20, 30, 40],
[15, 25, 35, 45],
[27, 29, 37, 48],
[32, 33, 39, 50]]
printf(mat)
|
C#
using System;
class GFG {
static void print(int[, ] mat)
{
for (int i = 0; i < mat.GetLength(0); i++) {
if (i % 2 == 0) {
for (int j = 0; j < mat.GetLength(1); j++)
Console.Write(mat[i, j] + " ");
}
else {
for (int j = mat.GetLength(1) - 1; j >= 0;
j--)
Console.Write(mat[i, j] + " ");
}
}
}
public static void Main()
{
int[, ] mat = { { 10, 20, 30, 40 },
{ 15, 25, 35, 45 },
{ 27, 29, 37, 48 },
{ 32, 33, 39, 50 } };
print(mat);
}
}
|
PHP
<?php
$M= 4;
$N =4;
function printLN($mat)
{
global $M;
global $N;
for ($i = 0; $i < $M; $i++)
{
if ($i % 2 == 0)
{
for ($j = 0; $j < $N; $j++)
echo $mat[$i][$j], " ";
}
else
{
for ($j = $N - 1; $j >= 0; $j--)
echo $mat[$i][$j] , " ";
}
}
}
$mat = array(array(10, 20, 30, 40),
array(15, 25, 35, 45),
array(27, 29, 37, 48),
array(32, 33, 39, 50));
printLN($mat);
?>
|
Javascript
<script>
let M = 4;
let N = 4;
function print(mat)
{
for(let i = 0; i < M; i++)
{
if (i % 2 == 0)
{
for(let j = 0; j < N; j++)
document.write(mat[i][j] + " ");
}
else
{
for(let j = N - 1; j >= 0; j--)
document.write(mat[i][j] + " ");
}
}
}
let mat = [ [ 10, 20, 30, 40 ],
[ 15, 25, 35, 45 ],
[ 27, 29, 37, 48 ],
[ 32, 33, 39, 50 ] ];
print(mat);
</script>
|
Output
10 20 30 40 45 35 25 15 27 29 37 48 50 39 33 32
Time Complexity: O(N x M), Traversing over all the elements of the matrix, therefore N X M elements are there.
Auxiliary Space: O(1)
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