C++ Program for Mirror of matrix across diagonal

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++

// 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;
}

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)

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++

// 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;
}

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

Please refer complete article on Mirror of matrix across diagonal for more details!