Mirror of matrix across diagonal

15 Jun, 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. 
C++
Java
Python3
C#
Javascript
// Simple CPP program to find mirror of// matrix across diagonal.#include <bits/stdc++.h>using namespace std; const int MAX = 100; void imageSwap(int mat[][MAX], 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<int> s;        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.top();            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<int> s;        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.top();            s.pop();        }    }} // Utility function to print a matrixvoid printMatrix(int mat[][MAX], int n){    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++)            cout << mat[i][j] << " ";        cout << endl;    }} // driver program to test above functionint main(){    int mat[][MAX] = { { 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);    return 0;}
// 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
# Simple Python3 program to find mirror of# matrix across diagonal.MAX = 100 def imageSwap(mat, n):         # for diagonal which start from at    # first row of matrix    row = 0         # traverse all top right diagonal    for j in range(n):                 # here we use stack for reversing        # the element of diagonal        s = []        i = row        k = j        while (i < n and k >= 0):            s.append(mat[i][k])            i += 1            k -= 1                     # push all element back to matrix        # in reverse order        i = row        k = j        while (i < n and k >= 0):            mat[i][k] = s[-1]            k -= 1            i += 1            s.pop()                 # do the same process for all the    # diagonal which start from last    # column    column = n - 1    for j in range(1, n):                 # here we use stack for reversing        # the elements of diagonal        s = []        i = j        k = column        while (i < n and k >= 0):            s.append(mat[i][k])            i += 1            k -= 1                     # push all element back to matrix        # in reverse order        i = j        k = column        while (i < n and k >= 0):            mat[i][k] = s[-1]            i += 1            k -= 1            s.pop() # Utility function to print a matrixdef printMatrix(mat, n):    for i in range(n):        for j in range(n):            print(mat[i][j], end=" ")        print()         # Driver codemat = [[1, 2, 3, 4],[5, 6, 7, 8],        [9, 10, 11, 12],[13, 14, 15, 16]]n = 4imageSwap(mat, n)printMatrix(mat, n) # This code is contributed by shubhamsingh10
// Simple C# program to find mirror of// matrix across diagonal.using System;using System.Collections.Generic; 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<int> s = new Stack<int>();            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<int> s = new Stack<int>();            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++)            {                Console.Write(mat[i,j] + " ");            }            Console.WriteLine("");        }    }     // Driver code    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 PrinciRaj1992 */
<script>       // Simple Javascript program to find mirror of matrix across diagonal.         let MAX = 100;      function imageSwap(mat, n)    {        // for diagonal which start from at        // first row of matrix        let row = 0;          // traverse all top right diagonal        for (let j = 0; j < n; j++)        {              // here we use stack for reversing            // the element of diagonal            let s = [];            let 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[s.length - 1];                s.pop();            }        }          // do the same process for all the        // diagonal which start from last        // column        let column = n - 1;        for (let j = 1; j < n; j++)        {              // here we use stack for reversing            // the elements of diagonal            let s = [];            let 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[s.length - 1];                s.pop();            }        }    }      // Utility function to print a matrix    function printMatrix(mat, n)    {        for (let i = 0; i < n; i++)        {            for (let j = 0; j < n; j++)            {                document.write(mat[i][j] + " ");            }            document.write("</br>");        }    }         let mat = [[1, 2, 3, 4],               [5, 6, 7, 8],               [9, 10, 11, 12],               [13, 14, 15, 16]];    let n = 4;    imageSwap(mat, n);    printMatrix(mat, n);     </script>
Output: 
1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16
Time complexity : O(n2)
Auxiliary Space: O(n), as stack is used 
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. 
C++
Java
Python3
C#
PHP
Javascript
// Efficient CPP program to find mirror of// matrix across diagonal.#include <bits/stdc++.h>using namespace std; const int MAX = 100; void imageSwap(int mat[][MAX], 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 matrixvoid printMatrix(int mat[][MAX], int n){    for (int i = 0; i < n; i++) {        for (int j = 0; j < n; j++)            cout << mat[i][j] << " ";        cout << endl;    }} // driver program to test above functionint main(){    int mat[][MAX] = { { 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);    return 0;}
// 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.
# Efficient Python3 program to find mirror of# matrix across diagonal.from builtins import rangeMAX = 100; def imageSwap(mat, n):     # traverse a matrix and swap    # mat[i][j] with mat[j][i]    for i in range(n):        for j in range(i + 1):            t = mat[i][j];            mat[i][j] = mat[j][i]            mat[j][i] = t # Utility function to print a matrixdef printMatrix(mat, n):    for i in range(n):        for j in range(n):            print(mat[i][j], end=" ");        print(); # Driver codeif __name__ == '__main__':    mat = [1, 2, 3, 4], \        [5, 6, 7, 8], \        [9, 10, 11, 12], \        [13, 14, 15, 16];    n = 4;    imageSwap(mat, n);    printMatrix(mat, n); # This code is contributed by Rajput-Ji
// Efficient C# program to find mirror of// matrix across diagonal.using System;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++)                Console.Write( mat[i,j] + " ");            Console.WriteLine();        }    }         // driver program to test above function    public static void Main ()    {        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.
<?php// Efficient PHP program to find mirror// of matrix across diagonal. function imageSwap(&$mat, $n){    // traverse a matrix and swap    // mat[i][j] with mat[j][i]    for ($i = 0; $i < $n; $i++)        for ($j = 0; $j <= $i; $j++)            $mat[$i][$j] = $mat[$i][$j] + $mat[$j][$i] -                          ($mat[$j][$i] = $mat[$i][$j]);} // Utility function to print a matrixfunction printMatrix(&$mat, $n){    for ($i = 0; $i < $n; $i++)    {        for ($j = 0; $j < $n; $j++)        {            echo ($mat[$i][$j]);            echo (" ");        }        echo ("\n");    }} // Driver Code$mat = array(array(1, 2, 3, 4),             array(5, 6, 7, 8),             array(9, 10, 11, 12),             array(13, 14, 15, 16));$n = 4;imageSwap($mat, $n);printMatrix($mat, $n); // This code is contributed// by Shivi_Aggarwal?>
<script>    // Efficient Javascript program to find mirror of    // matrix across diagonal.         let MAX = 100;           function imageSwap(mat, n)    {                   // traverse a matrix and swap        // mat[i][j] with mat[j][i]        for (let i = 0; i < n; i++)            for (let 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    function printMatrix(mat, n)    {        for (let i = 0; i < n; i++) {            for (let j = 0; j < n; j++)                document.write(mat[i][j] + " ");            document.write("</br>");        }    }         let mat = [ [ 1, 2, 3, 4 ],                [ 5, 6, 7, 8 ],                [ 9, 10, 11, 12 ],                [ 13, 14, 15, 16 ] ];    let n = 4;    imageSwap(mat, n);    printMatrix(mat, n);       </script>
Output: 
1 5 9 13 
2 6 10 14 
3 7 11 15 
4 8 12 16 
Time complexity : O(n2)Auxiliary Space: O(1)
This article is contributed by Aarti_Rathi. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above. 
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