Program to Interchange Diagonals of Matrix
Given a square matrix of order n*n, you have to interchange the elements of both diagonals.
Examples :
Input : matrix[][] = {1, 2, 3, 4, 5, 6, 7, 8, 9} Output : matrix[][] = {3, 2, 1, 4, 5, 6, 9, 8, 7} Input : matrix[][] = {4, 2, 3, 1, 5, 7, 6, 8, 9, 11, 10, 12, 16, 14, 15, 13} Output : matrix[][] = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 11, 14, 15, 16}
Explanation : Idea behind interchanging diagonals of a square matrix is simple. Iterate from 0 to n-1 and for each iteration you have to swap a[i][i] and a[i][n-i-1].
Implementation:
C++
// C++ program to interchange // the diagonals of matrix #include<bits/stdc++.h> using namespace std; #define N 3 // Function to interchange diagonals void interchangeDiagonals( int array[][N]) { // swap elements of diagonal for ( int i = 0; i < N; ++i) if (i != N / 2) swap(array[i][i], array[i][N - i - 1]); for ( int i = 0; i < N; ++i) { for ( int j = 0; j < N; ++j) cout<< " " << array[i][j]; cout<<endl; } } // Driver Code int main() { int array[N][N] = {4, 5, 6, 1, 2, 3, 7, 8, 9}; interchangeDiagonals(array); return 0; } // This code is contributed by noob2000. |
C
// C program to interchange // the diagonals of matrix #include<bits/stdc++.h> using namespace std; #define N 3 // Function to interchange diagonals void interchangeDiagonals( int array[][N]) { // swap elements of diagonal for ( int i = 0; i < N; ++i) if (i != N / 2) swap(array[i][i], array[i][N - i - 1]); for ( int i = 0; i < N; ++i) { for ( int j = 0; j < N; ++j) printf ( " %d" , array[i][j]); printf ( "\n" ); } } // Driver Code int main() { int array[N][N] = {4, 5, 6, 1, 2, 3, 7, 8, 9}; interchangeDiagonals(array); return 0; } |
Java
// Java program to interchange // the diagonals of matrix import java.io.*; class GFG { public static int N = 3 ; // Function to interchange diagonals static void interchangeDiagonals( int array[][]) { // swap elements of diagonal for ( int i = 0 ; i < N; ++i) if (i != N / 2 ) { int temp = array[i][i]; array[i][i] = array[i][N - i - 1 ]; array[i][N - i - 1 ] = temp; } for ( int i = 0 ; i < N; ++i) { for ( int j = 0 ; j < N; ++j) System.out.print(array[i][j]+ " " ); System.out.println(); } } // Driver Code public static void main (String[] args) { int array[][] = { { 4 , 5 , 6 }, { 1 , 2 , 3 }, { 7 , 8 , 9 } }; interchangeDiagonals(array); } } // This code is contributed by Pramod Kumar |
Python3
# Python program to interchange # the diagonals of matrix N = 3 ; # Function to interchange diagonals def interchangeDiagonals(array): # swap elements of diagonal for i in range (N): if (i ! = N / 2 ): temp = array[i][i]; array[i][i] = array[i][N - i - 1 ]; array[i][N - i - 1 ] = temp; for i in range (N): for j in range (N): print (array[i][j], end = " " ); print (); # Driver Code if __name__ = = '__main__' : array = [ 4 , 5 , 6 ],[ 1 , 2 , 3 ],[ 7 , 8 , 9 ]; interchangeDiagonals(array); # This code is contributed by Rajput-Ji |
C#
// C# program to interchange // the diagonals of matrix using System; class GFG { public static int N = 3; // Function to interchange diagonals static void interchangeDiagonals( int [,]array) { // swap elements of diagonal for ( int i = 0; i < N; ++i) if (i != N / 2) { int temp = array[i, i]; array[i, i] = array[i, N - i - 1]; array[i, N - i - 1] = temp; } for ( int i = 0; i < N; ++i) { for ( int j = 0; j < N; ++j) Console.Write(array[i, j]+ " " ); Console.WriteLine(); } } // Driver Code public static void Main () { int [,]array = { {4, 5, 6}, {1, 2, 3}, {7, 8, 9} }; interchangeDiagonals(array); } } // This code is contributed by vt_m. |
Javascript
<script> // Javascript program to interchange // the diagonals of matrix let N = 3; // Function to interchange diagonals function interchangeDiagonals(array) { // swap elements of diagonal for (let i = 0; i < N; ++i) if (i != parseInt(N / 2)) { let temp = array[i][i]; array[i][i] = array[i][N - i - 1]; array[i][N - i - 1] = temp; } for (let i = 0; i < N; ++i) { for (let j = 0; j < N; ++j) document.write( " " + array[i][j]); document.write( "<br>" ); } } // Driver Code let array = [[4, 5, 6], [1, 2, 3], [7, 8, 9]]; interchangeDiagonals(array); // This code is contributed by subham348. </script> |
Output
6 5 4 1 2 3 9 8 7
Time Complexity: O(N*N), as we are using nested loops for traversing the matrix.
Auxiliary Space: O(1), as we are not using any extra space.
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