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Java Program for Mirror of matrix across diagonal

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  • Last Updated : 12 Sep, 2022
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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




// Simple Java program to find mirror of
// matrix across diagonal.
 
import java.util.*;
 
class GFG
{
 
    static int MAX = 100;
 
    static void imageSwap(int mat[][], int n)
    {
        // for diagonal which start from at
        // first row of matrix
        int row = 0;
 
        // traverse all top right diagonal
        for (int j = 0; j < n; j++)
        {
 
            // here we use stack for reversing
            // the element of diagonal
            Stack<Integer> s = new Stack<>();
            int i = row, k = j;
            while (i < n && k >= 0)
            {
                s.push(mat[i++][k--]);
            }
 
            // push all element back to matrix
            // in reverse order
            i = row;
            k = j;
            while (i < n && k >= 0)
            {
                mat[i++][k--] = s.peek();
                s.pop();
            }
        }
 
        // do the same process for all the
        // diagonal which start from last
        // column
        int column = n - 1;
        for (int j = 1; j < n; j++)
        {
 
            // here we use stack for reversing
            // the elements of diagonal
            Stack<Integer> s = new Stack<>();
            int i = j, k = column;
            while (i < n && k >= 0)
            {
                s.push(mat[i++][k--]);
            }
 
            // push all element back to matrix
            // in reverse order
            i = j;
            k = column;
            while (i < n && k >= 0)
            {
                mat[i++][k--] = s.peek();
                s.pop();
            }
        }
    }
 
    // Utility function to print a matrix
    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("");
        }
    }
 
    // Driver program to test above function
    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);
    }
}
 
// This code contributed by Rajput-Ji


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




// Efficient Java program to find mirror of
// matrix across diagonal.
import java.io.*;
 
class GFG {
     
    static int MAX = 100;
     
    static void imageSwap(int mat[][], int n)
    {
         
        // traverse a matrix and swap
        // mat[i][j] with mat[j][i]
        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]);    
    }
     
    // Utility function to print a matrix
    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();
        }
    }
     
    // driver program to test above function
    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);
    }
}
 
// This code is contributed by anuj_67.


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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