Find a Symmetric matrix of order N that contain integers from 0 to N-1 and main diagonal should contain only 0’s

Given an integer N. The task is to generate a symmetric matrix of order N*N having the following properties.

  1. Main diagonal should contain only 0’s
  2. The matrix should contain elements from 0 to N-1 only.

Examples:

Input: N = 4
Output:
0 2 3 1
2 0 1 3
3 1 0 2
1 3 2 0



Input: N = 5
Output:
0 2 3 4 1
2 0 4 1 3
3 4 0 2 1
4 1 2 0 3
1 3 1 3 0

Approach: Since the required matrix has to be a square matrix, we can generate a symmetric matrix containing an element from 1 to n-1, excluding 0. We will deal with the case of 0 later.
Take for example when N = 4:
We first generate a symmetric matrix, and it can be easily done by filling every row from 1 to n-1 in cyclic order, i.e. fill the first row by 1 2 3, and do this for all subsequent rows in cyclic order.

So, the final matrix will be,
1 2 3
2 3 1
3 1 2

Now, we have generated a symmetric matrix containing elements from 1 to n. Let’s discuss case 0. We will take the benefit of the above matrix being symmetrical, we will add a column of 0 and rows of 0 like this,

1 2 3 0
2 3 1 0
3 1 2 0
0 0 0 0

Now, we have to put all 0 in diagonal. For this, we will start with the first row till last-1 row and swap all the 0 with the number that is there in each row and will also make a change in the last row like this:

For row 1, we swap 0 and 1 and also put last row’s 1st element with the number we swapped i.e. 1.
0 2 3 1
2 3 1 0
3 1 2 0
1 0 0 0

For row 2 we swap 0 and 3, and make the second element of the last row also 3.
0 2 3 1
2 0 1 3
3 1 2 0
1 3 0 0
and so on…
The final matrix generated will be the required matrix.

Below is the implementation of the above approach:

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// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
  
// Function to generate the required matrix
void solve(long long n)
{
    long long initial_array[n - 1][n - 1], final_array[n][n];
  
    for (long long i = 0; i < n - 1; ++i)
        initial_array[0][i] = i + 1;
  
    // Form cyclic array of elements 1 to n-1
    for (long long i = 1; i < n - 1; ++i)
        for (long long j = 0; j < n - 1; ++j)
            initial_array[i][j]
                = initial_array[i - 1][(j + 1) % (n - 1)];
  
    // Store initial array into final array
    for (long long i = 0; i < n - 1; ++i)
        for (long long j = 0; j < n - 1; ++j)
            final_array[i][j] = initial_array[i][j];
  
    // Fill the last row and column with 0's
    for (long long i = 0; i < n; ++i)
        final_array[i][n - 1] = final_array[n - 1][i] = 0;
  
    for (long long i = 0; i < n; ++i) {
        long long t0 = final_array[i][i];
        long long t1 = final_array[i][n - 1];
  
        // Swap 0 and the number present
        // at the current indexed row
        swap(final_array[i][i], final_array[i][n - 1]);
  
        // Also make changes in the last row
        // with the number we swapped
        final_array[n - 1][i] = t0;
    }
  
    // Print the final array
    for (long long i = 0; i < n; ++i) {
        for (long long j = 0; j < n; ++j)
            cout << final_array[i][j] << " ";
        cout << endl;
    }
}
  
// Driver code
int main()
{
    long long n = 5;
    solve(n);
  
    return 0;
}
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// Java implementation of the approach 
class GFG
{
  
// Function to generate the required matrix 
static void solve(long n) 
    long initial_array[][]= new long[(int)n - 1][(int)n - 1], 
                    final_array[][]= new long[(int)n][(int)n]; 
  
    for (long i = 0; i < n - 1; ++i) 
        initial_array[0][(int)i] = i + 1
  
    // Form cyclic array of elements 1 to n-1 
    for (long i = 1; i < n - 1; ++i) 
        for (long j = 0; j < n - 1; ++j) 
            initial_array[(int)i][(int)j] 
                = initial_array[(int)i - 1][(int)((int)j + 1) % ((int)n - 1)]; 
  
    // Store initial array into final array 
    for (long i = 0; i < n - 1; ++i) 
        for (long j = 0; j < n - 1; ++j) 
            final_array[(int)i][(int)j] = initial_array[(int)i][(int)j]; 
  
    // Fill the last row and column with 0's 
    for (long i = 0; i < n; ++i) 
        final_array[(int)i][(int)n - 1] = final_array[(int)n - 1][(int)i] = 0
  
    for (long i = 0; i < n; ++i) 
    
        long t0 = final_array[(int)i][(int)i]; 
        long t1 = final_array[(int)i][(int)n - 1]; 
  
        // Swap 0 and the number present 
        // at the current indexed row 
        long s = final_array[(int)i][(int)i];
        final_array[(int)i][(int)i]=final_array[(int)i][(int)n - 1];
        final_array[(int)i][(int)n - 1]=s;
  
        // Also make changes in the last row 
        // with the number we swapped 
        final_array[(int)n - 1][(int)i] = t0; 
    
  
    // Print the final array 
    for (long i = 0; i < n; ++i) 
    
        for (long j = 0; j < n; ++j) 
            System.out.print( final_array[(int)i][(int)j] + " "); 
        System.out.println();
    
  
// Driver code 
public static void main(String args[])
    long n = 5
    solve(n); 
}
}
  
// This code is contributed by Arnab Kundu 
    
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# Python 3 implementation of the approach
  
# Function to generate the required matrix
def solve(n):
    initial_array = [[0 for i in range(n-1)] for j in range(n-1)]
    final_array = [[0 for i in range(n)]for j in range(n)]
  
    for i in range(n - 1):
        initial_array[0][i] = i + 1
  
    # Form cyclic array of elements 1 to n-1
    for i in range(1, n - 1):
        for j in range(n - 1):
            initial_array[i][j] = initial_array[i - 1][(j + 1) % (n - 1)]
  
    # Store initial array into final array
    for i in range(n-1):
        for j in range(n-1):
            final_array[i][j] = initial_array[i][j]
  
    # Fill the last row and column with 0's
    for i in range(n):
        final_array[i][n - 1] = final_array[n - 1][i] = 0
  
    for i in range(n):
        t0 = final_array[i][i]
        t1 = final_array[i][n - 1]
  
        # Swap 0 and the number present
        # at the current indexed row
        temp = final_array[i][i]
        final_array[i][i] = final_array[i][n - 1]
        final_array[i][n - 1] = temp
  
        # Also make changes in the last row
        # with the number we swapped
        final_array[n - 1][i] = t0
  
    # Print the final array
    for i in range(n):
        for j in range(n):
            print(final_array[i][j],end = " ")
        print("\n",end = "")
  
# Driver code
if __name__ == '__main__':
    n = 5
    solve(n)
      
# This code is contributed by
# Surendra_Gangwar
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// C# implementation of the approach 
using System;
  
class GFG
{
  
// Function to generate the required matrix 
static void solve(long n) 
    long [,]initial_array = new long[(int)n - 1,(int)n - 1]; 
    long [,]final_array = new long[(int)n,(int)n]; 
  
    for (long i = 0; i < n - 1; ++i) 
        initial_array[0,(int)i] = i + 1; 
  
    // Form cyclic array of elements 1 to n-1 
    for (long i = 1; i < n - 1; ++i) 
        for (long j = 0; j < n - 1; ++j) 
            initial_array[(int)i,(int)j] 
                = initial_array[(int)i - 1,(int)((int)j + 1) % ((int)n - 1)]; 
  
    // Store initial array into final array 
    for (long i = 0; i < n - 1; ++i) 
        for (long j = 0; j < n - 1; ++j) 
            final_array[(int)i,(int)j] = initial_array[(int)i,(int)j]; 
  
    // Fill the last row and column with 0's 
    for (long i = 0; i < n; ++i) 
        final_array[(int)i,(int)n - 1] = final_array[(int)n - 1,(int)i] = 0; 
  
    for (long i = 0; i < n; ++i) 
    
        long t0 = final_array[(int)i, (int)i]; 
        long t1 = final_array[(int)i, (int)n - 1]; 
  
        // Swap 0 and the number present 
        // at the current indexed row 
        long s = final_array[(int)i,(int)i];
        final_array[(int)i,(int)i] = final_array[(int)i, (int)n - 1];
        final_array[(int)i,(int)n - 1] = s;
  
        // Also make changes in the last row 
        // with the number we swapped 
        final_array[(int)n - 1,(int)i] = t0; 
    
  
    // Print the final array 
    for (long i = 0; i < n; ++i) 
    
        for (long j = 0; j < n; ++j) 
            Console.Write( final_array[(int)i,(int)j] + " "); 
        Console.WriteLine();
    
  
// Driver code 
public static void Main(String []args)
    long n = 5; 
    solve(n); 
}
}
  
// This code contributed by Rajput-Ji
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<?php
// Php implementation of the approach 
  
// Function to generate the required matrix 
function solve($n
    $initial_array = array(array()) ;
    $final_array = array(array()) ;
  
    for ($i = 0; $i < $n - 1; ++$i
        $initial_array[0][$i] = $i + 1; 
  
    // Form cyclic array of elements 1 to n-1 
    for ($i = 1; $i < $n - 1; ++$i
        for ($j = 0; $j < $n - 1; ++$j
            $initial_array[$i][$j] = 
                $initial_array[$i - 1][($j + 1) % ($n - 1)]; 
  
    // Store initial array into final array 
    for ($i = 0; $i < $n - 1; ++$i
        for ($j = 0; $j < $n - 1; ++$j
            $final_array[$i][$j] = $initial_array[$i][$j]; 
  
    // Fill the last row and column with 0's 
    for ($i = 0; $i < $n; ++$i
        $final_array[$i][$n - 1] = $final_array[$n - 1][$i] = 0; 
  
    for ($i = 0; $i < $n; ++$i)
    
        $t0 = $final_array[$i][$i]; 
        $t1 = $final_array[$i][$n - 1]; 
  
        // Swap 0 and the number present 
        // at the current indexed row 
        $temp = $final_array[$i][$i] ;
        $final_array[$i][$i] = $final_array[$i][$n - 1] ;
        $final_array[$i][$n - 1] = $temp ;
  
        // Also make changes in the last row 
        // with the number we swapped 
        $final_array[$n - 1][$i] = $t0
    
  
    // Print the final array 
    for ($i = 0; $i < $n; ++$i
    
        for ($j = 0; $j < $n; ++$j
            echo $final_array[$i][$j]," "
        echo "\n"
    
  
    // Driver code 
    $n = 5; 
    solve($n); 
      
// This code is contributed by Ryuga
?>
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Output:
0 2 3 4 1 
2 0 4 1 3 
3 4 0 2 1 
4 1 2 0 3 
1 3 1 3 0

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