Mirror of matrix across diagonal

Given a 2-D array of order N x N, print a matrix which is mirror of given tree across 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 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

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Simple solution of this problem consumes extra space, we traverse all right diagonal (right-to-left) one-by-one. During the traversal of diagonal, first we push all the elements into the stack and after we traverse it again and replace every element of diagonal with the stack element.

Below is the implementation of above idea.

C++

// 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 matrix 
void 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 function 
int 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; 
} 

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 

Python3

# 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 pra matrix 
def printMatrix(mat, n): 
    for i in range(n): 
        for j in range(n): 
            print(mat[i][j], end=" ") 
        print() 
          
# Driver code 
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 shubhamsingh10 

C#

// 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 */

Output:

Time complexity : O(n*n)

Efficient solution of this problem is that if we observe a output matrix than we notice that we just have to swap (mat[i][j] mat[j][i]).
Below is the implementation of above idea.\

C++

// 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 matrix 
void 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 function 
int 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; 
} 

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. 

Python3

# Efficient Python3 program to find mirror of 
# matrix across diagonal. 
from builtins import range
MAX = 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 pra matrix 
def printMatrix(mat, n): 
    for i in range(n): 
        for j in range(n): 
            print(mat[i][j], end=" "); 
        print(); 
  
# Driver code 
if __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 

C#

// 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

<?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 matrix 
function 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 
?> 

Output:

Time complexity : O(n*n)

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

Python3

C#

C++

Java

Python3

C#

PHP

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