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++
#include<bits/stdc++.h>
using namespace std;
#define N 3
void interchangeDiagonals( int array[][N])
{
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;
}
}
int main()
{
int array[N][N] = {4, 5, 6,
1, 2, 3,
7, 8, 9};
interchangeDiagonals(array);
return 0;
}
|
C
#include<bits/stdc++.h>
using namespace std;
#define N 3
void interchangeDiagonals( int array[][N])
{
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" );
}
}
int main()
{
int array[N][N] = {4, 5, 6,
1, 2, 3,
7, 8, 9};
interchangeDiagonals(array);
return 0;
}
|
Java
import java.io.*;
class GFG
{
public static int N = 3 ;
static void interchangeDiagonals( int array[][])
{
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();
}
}
public static void main (String[] args)
{
int array[][] = { { 4 , 5 , 6 },
{ 1 , 2 , 3 },
{ 7 , 8 , 9 }
};
interchangeDiagonals(array);
}
}
|
Python3
N = 3 ;
def interchangeDiagonals(array):
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 ();
if __name__ = = '__main__' :
array = [ 4 , 5 , 6 ],[ 1 , 2 , 3 ],[ 7 , 8 , 9 ];
interchangeDiagonals(array);
|
C#
using System;
class GFG
{
public static int N = 3;
static void interchangeDiagonals( int [,]array)
{
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();
}
}
public static void Main ()
{
int [,]array = { {4, 5, 6},
{1, 2, 3},
{7, 8, 9}
};
interchangeDiagonals(array);
}
}
|
Javascript
<script>
let N = 3;
function interchangeDiagonals(array)
{
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>" );
}
}
let array = [[4, 5, 6],
[1, 2, 3],
[7, 8, 9]];
interchangeDiagonals(array);
</script>
|
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.
Last Updated :
06 Oct, 2023
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