You are given a matrix of order N*N. The task is to find the resultant matrix by adding the mirror image of given matrix with the matrix itself.
Examples:
Input : mat[][] = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}}
Output : 4 4 4
10 10 10
16 16 16
Explanation:
Resultant Matrix = {{1, 2, 3}, {{3, 2, 1},
{4, 5, 6}, + {6, 5, 4},
{7, 8, 9}} {9, 8, 7}}
Input : mat[][] = {{1, 2},
{3, 4}}
Output : 3 3
7 7
While finding the mirror image of matrix the row of each element will remain same but the value of its columns will reshuffle. For any element Aij its new position in mirror image will be Ai(n-j). After getting the mirror image of matrix add it to original matrix and print the result.
Points to take care:
- Indexing of matrix will start from 0, 0 and ends on n-1, n-1 hence position of any element Aij will be Ai(n-1-j).
- While printing the result take care of proper output format
Below is the implementation of the above approach:
C++
// C++ program to find sum of matrix and// its mirror image#include <bits/stdc++.h>#define N 4using namespace std;// Function to print the resultant matrixvoid printSum(int mat[][N]){ for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { cout << setw(3) << mat[i][N - 1 - j] + mat[i][j] << " "; } cout << "\n"; }}// Driver Codeint main(){ int mat[N][N] = { { 2, 4, 6, 8 }, { 1, 3, 5, 7 }, { 8, 6, 4, 2 }, { 7, 5, 3, 1 } }; printSum(mat); return 0;} |
Java
// Java program to find sum of // matrix and its mirror imageimport java.io.*;class GFG {static int N = 4;// Function to print the // resultant matrixstatic void printSum(int mat[][]){ for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { System.out.print((mat[i][N - 1 - j] + mat[i][j]) + " "); } System.out.println(); }}// Driver Codepublic static void main (String[] args) { int mat[][] = { { 2, 4, 6, 8 }, { 1, 3, 5, 7 }, { 8, 6, 4, 2 }, { 7, 5, 3, 1 } }; printSum(mat);}}// This code is contributed by anuj_67 |
Python3
# Python 3 program to find sum of matrix # and its mirror imageN = 4# Function to print the resultant matrixdef printSum(mat): for i in range(N): for j in range(N): print('{:>3}'.format(mat[i][N - 1 - j] + mat[i][j]), end =" ") print("\n", end = "")# Driver Codeif __name__ == '__main__': mat = [[2, 4, 6, 8], [1, 3, 5, 7], [8, 6, 4, 2], [7, 5, 3, 1]] printSum(mat)# This code is contributed by# Surendra_Gangwar |
C#
// C# program to find sum of // matrix and its mirror imageusing System;class GFG {static int N = 4;// Function to print the // resultant matrixstatic void printSum(int [,]mat){ for (int i = 0; i < N; i++) { for (int j = 0; j < N; j++) { Console.Write((mat[i, N - 1 - j] + mat[i, j]) + " "); } Console.WriteLine(); }}// Driver Codepublic static void Main () { int [,]mat = { { 2, 4, 6, 8 }, { 1, 3, 5, 7 }, { 8, 6, 4, 2 }, { 7, 5, 3, 1 } }; printSum(mat);}}// This code is contributed by shs.. |
PHP
<?php// PHP program to find sum of // matrix and its mirror image// Function to print the // resultant matrixfunction printSum($mat){ for ($i = 0; $i < 4; $i++) { for ($j = 0; $j < 4; $j++) { echo ($mat[$i][4 - 1 - $j] + $mat[$i][$j]) . " " ; } echo "\n"; }}// Driver Code$mat = array(array(2, 4, 6, 8 ), array(1, 3, 5, 7), array(8, 6, 4, 2), array(7, 5, 3, 1));printSum($mat); // This code is contributed// by Akanksha Rai?> |
Javascript
<script>// Javascript program to find sum of matrix and// its mirror imagevar N = 4// Function to print the resultant matrixfunction printSum(mat){ for (var i = 0; i < N; i++) { for (var j = 0; j < N; j++) { document.write( (mat[i][N - 1 - j] + mat[i][j]) + " "); } document.write("<br>"); }}// Driver Codevar mat = [ [ 2, 4, 6, 8 ], [ 1, 3, 5, 7 ], [ 8, 6, 4, 2 ], [ 7, 5, 3, 1 ] ];printSum(mat);</script> |
Output
10 10 10 10 8 8 8 8 10 10 10 10 8 8 8 8
Complexity Analysis:
- Time complexity : O(N2) for given input matrix of size N*N
- Auxiliary Space: O(1)
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