Shift matrix elements row-wise by k
Last Updated :
20 Feb, 2023
Given a square matrix mat[][] and a number k. The task is to shift the first k elements of each row to the right of the matrix.
Examples :
Input : mat[N][N] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
k = 2
Output :mat[N][N] = {{3, 1, 2}
{6, 4, 5}
{9, 7, 8}}
Input : mat[N][N] = {{1, 2, 3, 4}
{5, 6, 7, 8}
{9, 10, 11, 12}
{13, 14, 15, 16}}
k = 2
Output :mat[N][N] = {{3, 4, 1, 2}
{7, 8, 5, 6}
{11, 12, 9, 10}
{15, 16, 13, 14}}
Note: Matrix should be a square matrix
Implementation:
C++
#include <bits/stdc++.h>
using namespace std;
#define N 4
void shiftMatrixByK( int mat[N][N], int k)
{
if (k > N) {
cout << "shifting is not possible" << endl;
return ;
}
int j = 0;
while (j < N) {
for ( int i = k; i < N; i++)
cout << mat[j][i] << " " ;
for ( int i = 0; i < k; i++)
cout << mat[j][i] << " " ;
cout << endl;
j++;
}
}
int main()
{
int mat[N][N] = {{1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16}};
int k = 2;
shiftMatrixByK(mat, k);
return 0;
}
|
Java
import java.io.*;
import java.util.*;
public class GFG {
static int N = 4 ;
static void shiftMatrixByK( int [][]mat,
int k)
{
if (k > N) {
System.out.print( "Shifting is"
+ " not possible" );
return ;
}
int j = 0 ;
while (j < N) {
for ( int i = k; i < N; i++)
System.out.print(mat[j][i] + " " );
for ( int i = 0 ; i < k; i++)
System.out.print(mat[j][i] + " " );
System.out.println();
j++;
}
}
public static void main(String args[])
{
int [][]mat = new int [][]
{ { 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 9 , 10 , 11 , 12 },
{ 13 , 14 , 15 , 16 } };
int k = 2 ;
shiftMatrixByK(mat, k);
}
}
|
Python3
N = 4
def shiftMatrixByK(mat, k):
if (k > N) :
print ( "shifting is"
" not possible" )
return
j = 0
while (j < N) :
for i in range (k, N):
print ( "{} " .
format (mat[j][i]), end = "")
for i in range ( 0 , k):
print ( "{} " .
format (mat[j][i]), end = "")
print ("")
j = j + 1
mat = [[ 1 , 2 , 3 , 4 ],
[ 5 , 6 , 7 , 8 ],
[ 9 , 10 , 11 , 12 ],
[ 13 , 14 , 15 , 16 ]]
k = 2
shiftMatrixByK(mat, k)
|
C#
using System;
class GFG {
static int N = 4;
static void shiftMatrixByK( int [,]mat,
int k)
{
if (k > N) {
Console.WriteLine( "shifting is"
+ " not possible" );
return ;
}
int j = 0;
while (j < N) {
for ( int i = k; i < N; i++)
Console.Write(mat[j,i] + " " );
for ( int i = 0; i < k; i++)
Console.Write(mat[j,i] + " " );
Console.WriteLine();
j++;
}
}
public static void Main()
{
int [,]mat = new int [,]
{ {1, 2, 3, 4},
{5, 6, 7, 8},
{9, 10, 11, 12},
{13, 14, 15, 16} };
int k = 2;
shiftMatrixByK(mat, k);
}
}
|
PHP
<?php
function shiftMatrixByK( $mat , $k )
{
$N = 4;
if ( $k > $N )
{
echo ( "shifting is not possible\n" );
return ;
}
$j = 0;
while ( $j < $N )
{
for ( $i = $k ; $i < $N ; $i ++)
echo ( $mat [ $j ][ $i ]. " " );
for ( $i = 0; $i < $k ; $i ++)
echo ( $mat [ $j ][ $i ]. " " );
echo ( "\n" );
$j ++;
}
}
$mat = array ( array (1, 2, 3, 4),
array (5, 6, 7, 8),
array (9, 10, 11, 12),
array (13, 14, 15, 16));
$k = 2;
shiftMatrixByK( $mat , $k );
?>
|
Javascript
<script>
let N = 4;
function shiftMatrixByK(mat,k)
{
if (k > N) {
document.write( "Shifting is not possible" );
return ;
}
let j = 0;
while (j < N) {
for (let i = k; i < N; i++)
document.write(mat[j][i] + " " );
for (let i = 0; i < k; i++)
document.write(mat[j][i] + " " );
document.write( "<br>" );
j++;
}
}
let mat =
[[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]];
let k = 2;
shiftMatrixByK(mat, k);
</script>
|
Output
3 4 1 2
7 8 5 6
11 12 9 10
15 16 13 14
Complexity Analysis:
- Time Complexity: O(n2),
- Auxiliary Space: O(1)
Like Article
Suggest improvement
Share your thoughts in the comments
Please Login to comment...