Java Program for Mirror of matrix across diagonal
Last Updated :
12 Sep, 2022
Given a 2-D array of order N x N, print a matrix that is the mirror of the given tree across the diagonal. We need to print the result in a way: swap the values of the triangle above the diagonal with the values of the triangle below it like a mirror image swap. Print the 2-D array obtained in a matrix layout.
Examples:
Input : int mat[][] = {{1 2 4 }
{5 9 0}
{ 3 1 7}}
Output : 1 5 3
2 9 1
4 0 7
Input : mat[][] = {{1 2 3 4 }
{5 6 7 8 }
{9 10 11 12}
{13 14 15 16} }
Output : 1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
A simple solution to this problem involves extra space. We traverse all right diagonal (right-to-left) one by one. During the traversal of the diagonal, first, we push all the elements into the stack and then we traverse it again and replace every element of the diagonal with the stack element.
Below is the implementation of the above idea.
Java
import java.util.*;
class GFG
{
static int MAX = 100 ;
static void imageSwap( int mat[][], int n)
{
int row = 0 ;
for ( int j = 0 ; j < n; j++)
{
Stack<Integer> s = new Stack<>();
int i = row, k = j;
while (i < n && k >= 0 )
{
s.push(mat[i++][k--]);
}
i = row;
k = j;
while (i < n && k >= 0 )
{
mat[i++][k--] = s.peek();
s.pop();
}
}
int column = n - 1 ;
for ( int j = 1 ; j < n; j++)
{
Stack<Integer> s = new Stack<>();
int i = j, k = column;
while (i < n && k >= 0 )
{
s.push(mat[i++][k--]);
}
i = j;
k = column;
while (i < n && k >= 0 )
{
mat[i++][k--] = s.peek();
s.pop();
}
}
}
static void printMatrix( int mat[][], int n)
{
for ( int i = 0 ; i < n; i++)
{
for ( int j = 0 ; j < n; j++)
{
System.out.print(mat[i][j] + " " );
}
System.out.println( "" );
}
}
public static void main(String[] args)
{
int mat[][] = {{ 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 9 , 10 , 11 , 12 },
{ 13 , 14 , 15 , 16 }};
int n = 4 ;
imageSwap(mat, n);
printMatrix(mat, n);
}
}
|
Output:
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
Time complexity : O(n*n)
An efficient solution to this problem is that if we observe an output matrix, then we notice that we just have to swap (mat[i][j] to mat[j][i]).
Below is the implementation of the above idea.
Java
import java.io.*;
class GFG {
static int MAX = 100 ;
static void imageSwap( int mat[][], int n)
{
for ( int i = 0 ; i < n; i++)
for ( int j = 0 ; j <= i; j++)
mat[i][j] = mat[i][j] + mat[j][i]
- (mat[j][i] = mat[i][j]);
}
static void printMatrix( int mat[][], int n)
{
for ( int i = 0 ; i < n; i++) {
for ( int j = 0 ; j < n; j++)
System.out.print( mat[i][j] + " " );
System.out.println();
}
}
public static void main (String[] args)
{
int mat[][] = { { 1 , 2 , 3 , 4 },
{ 5 , 6 , 7 , 8 },
{ 9 , 10 , 11 , 12 },
{ 13 , 14 , 15 , 16 } };
int n = 4 ;
imageSwap(mat, n);
printMatrix(mat, n);
}
}
|
Output:
1 5 9 13
2 6 10 14
3 7 11 15
4 8 12 16
Time complexity : O(n*n)
Auxiliary space: O(1) as it is using constant space for variables
Please refer complete article on Mirror of matrix across diagonal for more details!
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