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Construct a square Matrix whose parity of diagonal sum is same as size of matrix

  • Last Updated : 08 Apr, 2021

Given an integer N representing the size of the matrix, the task is to construct a square matrix N * N which have an element from 1 to N2 such that the parity of the sum of its diagonals is equal to the parity of integer N.

Examples:

Input: N = 4
Output:
1   2   3   4 
8   7   6   5
9  10  11 12
16 15 14 13
Explanation:
Sum of diagonal = 32 and 36 and integer N = 4, all the numbers are even that is same parity.

Input: N = 3
Output:
1 2 3 
6 5 4
7 8 9
Explanation:
Sum of diagonal = 15 and integer N = 3, all the numbers are odd that is same parity.

Approach: The idea is to observe that on filling the elements in the matrix in an alternative fashion the parity of N and the sum of diagonals is the same. Start the counter from 1 and then fill the first row from 0 to N – 1 in increasing order, then fill the second row from index N – 1 to 0, and so on. Keep filling each element from value 1 to N2 in this alternate fashion to get the required matrix.
Below is the implementation of the above approach:
 



C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to construct a N * N
// matrix based on the given condition
void createMatrix(int N)
{
    // Matrix with given sizM
    int M[N][N];
 
    // Counter to insert elements
    // from 1 to N * N
    int count = 1;
    for (int i = 0; i < N; i++) {
 
        // Check if it is first row
        // or odd row of the matrix
        if (i % 2 == 0) {
 
            for (int j = 0; j < N; j++) {
                M[i][j] = count;
                count++;
            }
        }
        else
        {
            // Insert elements from
            // right to left
            for (int j = N - 1;j >= 0; j--){
                M[i][j] = count;
                count += 1;
            }
        }
    }
 
    // Print the matrix
    for (int i = 0; i < N; i++) {
 
        // Traverse column
        for (int j = 0; j < N; j++) {
            cout << M[i][j] << " ";
        }
        cout << endl;
    }
}
 
// Driver Code
int main()
{
    // Given size of matrix N
    int N = 3;
 
    // Function Call
    createMatrix(N);
    return 0;
}

Java




// Java program for the above approach
class GFG{
  
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
    // Matrix with given sizM
    int M[][] = new int[N][N];
  
    // Counter to insert elements
    // from 1 to N * N
    int count = 1;
    for (int i = 0; i < N; i++)
    {
  
        // Check if it is first row
        // or odd row of the matrix
        if (i % 2 == 0)
        {
  
            for (int j = 0; j < N; j++)
            {
                M[i][j] = count;
                count++;
            }
        }
        else
        {
            // Insert elements from
            // right to left
            for(int j = N - 1; j >= 0; j--){
                M[i][j] = count;
                count += 1 ;
            }
        }
    }
  
    // Print the matrix
    for (int i = 0; i < N; i++)
    {
  
        // Traverse column
        for (int j = 0; j < N; j++)
        {
            System.out.print(M[i][j] + " ");
        }
        System.out.println();
    }
}
  
// Driver Code
public static void main(String[] args)
{
    // Given size of matrix N
    int N = 3;
  
    // Function Call
    createMatrix(N);
}
}
 
// This code is contributed by Ritik Bansal

Python3




# Python3 program for the above approach
 
# Function to construct a N * N
# matrix based on the given condition
def createMatrix(N):
     
    # Matrix with given size
    M = [[0] * N for i in range(N)]
 
    # Counter to insert elements
    # from 1 to N * N
    count = 1
     
    for i in range(N):
 
        # Check if it is first row
        # or odd row of the matrix
        if (i % 2 == 0):
            for j in range(N):
 
                # Insert elements from
                # left to right
                M[i][j] = count
                count += 1
 
        # Condition if it is second
        # row or even row
        else:
 
            # Insert elements from
            # right to left
            for j in range(N - 1, -1, -1):
                M[i][j] = count
                count += 1
 
    # Print the matrix
    for i in range(N):
 
        # Traverse column
        for j in range(N):
            print(M[i][j], end = " ")
 
        print()
 
# Driver Code
if __name__ == '__main__':
     
    # Given size of matrix N
    N = 3
 
    # Function call
    createMatrix(N)
 
# This code is contributed by mohit kumar 29

C#




// C# program for
// the above approach
using System;
class GFG{
  
// Function to construct a N * N
// matrix based on the given condition
static void createMatrix(int N)
{
  // Matrix with given sizM
  int[,] M = new int[N, N];
 
  // Counter to insert elements
  // from 1 to N * N
  int count = 1;
  for (int i = 0; i < N; i++)
  {
    // Check if it is first row
    // or odd row of the matrix
    if (i % 2 == 0)
    {
 
      for (int j = 0; j < N; j++)
      {
        M[i, j] = count;
        count++;
      }
    }
    else
    {
      // Insert elements from
      // right to left
      for(int j = N - 1; j >= 0; j--)
      {
        M[i, j] = count;
        count += 1;
      }
    }
  }
 
  // Print the matrix
  for (int i = 0; i < N; i++)
  {
    // Traverse column
    for (int j = 0; j < N; j++)
    {
      Console.Write(M[i, j] + " ");
    }
    Console.WriteLine();
  }
}
  
// Driver Code
public static void Main()
{
  // Given size of matrix N
  int N = 3;
 
  // Function Call
  createMatrix(N);
}
}
 
// This code is contributed by Chitranayal

Javascript




<script>
 
// javascript program for the above approach
 
  
// Function to construct a N * N
// matrix based on the given condition
function createMatrix(N)
{
    // Matrix with given sizM
    var M = Array(N).fill(0).map(x => Array(N).fill(0));
  
    // Counter to insert elements
    // from 1 to N * N
    var count = 1;
    for (i = 0; i < N; i++)
    {
  
        // Check if it is first row
        // or odd row of the matrix
        if (i % 2 == 0)
        {
  
            for (j = 0; j < N; j++)
            {
                M[i][j] = count;
                count++;
            }
        }
        else
        {
            // Insert elements from
            // right to left
            for(j = N - 1; j >= 0; j--){
                M[i][j] = count;
                count += 1 ;
            }
        }
    }
  
    // Prvar the matrix
    for (i = 0; i < N; i++)
    {
  
        // Traverse column
        for (j = 0; j < N; j++)
        {
            document.write(M[i][j] + " ");
        }
        document.write("<br>");
    }
}
  
// Driver Code
var N = 3;
 
// Function Call
createMatrix(N);
 
// This code is contributed by 29AjayKumar
 
</script>
Output: 
1 2 3 
6 5 4 
7 8 9

 

Time Complexity: O(N*N) 
Auxiliary Space: O(1)

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