Print sum of matrix and its mirror image
Last Updated :
09 Sep, 2022
You are given a matrix of order N*N. The task is to find the resultant matrix by adding the mirror image of given matrix with the matrix itself.
Examples:
Input : mat[][] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
Output : 4 4 4
10 10 10
16 16 16
Explanation:
Resultant Matrix = {{1, 2, 3}, {{3, 2, 1},
{4, 5, 6}, + {6, 5, 4},
{7, 8, 9}} {9, 8, 7}}
Input : mat[][] = {{1, 2},
{3, 4}}
Output : 3 3
7 7
While finding the mirror image of matrix the row of each element will remain same but the value of its columns will reshuffle. For any element Aij its new position in mirror image will be Ai(n-j). After getting the mirror image of matrix add it to original matrix and print the result.
Points to take care:
- Indexing of matrix will start from 0, 0 and ends on n-1, n-1 hence position of any element Aij will be Ai(n-1-j).
- While printing the result take care of proper output format
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
#define N 4
using namespace std;
void printSum( int mat[][N])
{
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++) {
cout << setw(3) << mat[i][N - 1 - j] + mat[i][j] << " " ;
}
cout << "\n" ;
}
}
int main()
{
int mat[N][N] = { { 2, 4, 6, 8 },
{ 1, 3, 5, 7 },
{ 8, 6, 4, 2 },
{ 7, 5, 3, 1 } };
printSum(mat);
return 0;
}
|
Java
import java.io.*;
class GFG
{
static int N = 4 ;
static void printSum( int mat[][])
{
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < N; j++)
{
System.out.print((mat[i][N - 1 - j] +
mat[i][j]) + " " );
}
System.out.println();
}
}
public static void main (String[] args)
{
int mat[][] = { { 2 , 4 , 6 , 8 },
{ 1 , 3 , 5 , 7 },
{ 8 , 6 , 4 , 2 },
{ 7 , 5 , 3 , 1 } };
printSum(mat);
}
}
|
Python3
N = 4
def printSum(mat):
for i in range (N):
for j in range (N):
print ( '{:>3}' . format (mat[i][N - 1 - j] +
mat[i][j]), end = " " )
print ( "\n" , end = "")
if __name__ = = '__main__' :
mat = [[ 2 , 4 , 6 , 8 ],
[ 1 , 3 , 5 , 7 ],
[ 8 , 6 , 4 , 2 ],
[ 7 , 5 , 3 , 1 ]]
printSum(mat)
|
C#
using System;
class GFG
{
static int N = 4;
static void printSum( int [,]mat)
{
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
Console.Write((mat[i, N - 1 - j] +
mat[i, j]) + " " );
}
Console.WriteLine();
}
}
public static void Main ()
{
int [,]mat = { { 2, 4, 6, 8 },
{ 1, 3, 5, 7 },
{ 8, 6, 4, 2 },
{ 7, 5, 3, 1 } };
printSum(mat);
}
}
|
PHP
<?php
function printSum( $mat )
{
for ( $i = 0; $i < 4; $i ++)
{
for ( $j = 0; $j < 4; $j ++)
{
echo ( $mat [ $i ][4 - 1 - $j ] +
$mat [ $i ][ $j ]) . " " ;
}
echo "\n" ;
}
}
$mat = array ( array (2, 4, 6, 8 ),
array (1, 3, 5, 7),
array (8, 6, 4, 2),
array (7, 5, 3, 1));
printSum( $mat );
?>
|
Javascript
<script>
var N = 4
function printSum(mat)
{
for ( var i = 0; i < N; i++) {
for ( var j = 0; j < N; j++) {
document.write( (mat[i][N - 1 - j] + mat[i][j]) + " " );
}
document.write( "<br>" );
}
}
var mat = [ [ 2, 4, 6, 8 ],
[ 1, 3, 5, 7 ],
[ 8, 6, 4, 2 ],
[ 7, 5, 3, 1 ] ];
printSum(mat);
</script>
|
Output
10 10 10 10
8 8 8 8
10 10 10 10
8 8 8 8
Complexity Analysis:
- Time complexity : O(N2) for given input matrix of size N*N
- Auxiliary Space: O(1)
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