Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Construct a Matrix N x N with first N^2 natural numbers for an input N

  • Difficulty Level : Medium
  • Last Updated : 26 Apr, 2021

Given an integer N, the task is to construct a matrix M[][] of size N x N with numbers in the range [1, N^2] with the following conditions :

  1. The elements of the matrix M should be an integer between 1 and N^2.
  2. All elements of the matrix M are pairwise distinct.
  3. For each square submatrix containing cells in row r through r+a and in columns c through c+a(inclusive) for some valid integers r,c and a>=0: M(r,c)+M(r+a,c+a) is even and M(r,c+a)+M(r+a,c) is even.
     

Examples: 
 

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

Input: N = 2 
Output: 
1 2 
4 3 
Explanation: 
This matrix has 5 square submatrix and 4 of them ([1], [2], [3], [4]) have a=0 so they satisfy the conditions. 
The last square submatrix is the whole matrix M where r=c=a=1. We can see that M(1, 1)+M(2, 2)=1+3=4 and M(1, 2)+M(2, 1)=2+4=6 are both even.
Input: N = 4 
Output: 
1 2 3 4 
8 7 6 5 
9 10 11 12 
16 15 14 13 
 



 

 

Approach: We know that the sum of two numbers is even when their parity is the same. Let us say the parity of M(i, j) is odd that means the parity of M(i+1, j+1), M(i+1, j-1), M(i-1, j+1), M(i-1, j-1) has to be odd.
Below is the illustration for N = 4 to generate a matrix of size 4×4: 
 

 

 

So from the above illustration we have to fill the matrix in the Checkerboard Pattern. We can fill it in two ways: 
 

 

  • All black cells have an odd integer and white cells have an even integer.
  • All black cells have an even integer and white cells have an odd integer.

 

Below is the implementation of the above approach:
  

C++




// C++ program for the above approach
#include<bits/stdc++.h>
using namespace std;
 
// Function to print the desired matrix
void UniqueMatrix(int N)
{
    int element_value = 1;
    int i = 0;
 
    // element_value will start from 1
    // and go up to N ^ 2
 
    // i is row number and it starts
    // from 0 and go up to N-1
     
    // Iterate ove all [0, N]
    while(i < N)
    {
         
        // If is even
        if(i % 2 == 0)
        {
            for(int f = element_value;
                    f < element_value + N; f++)
            {
                 
                // If row number is even print
                // the row in forward order
                cout << f << " ";
            }
            element_value += N;
        }
        else
        {
            for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
            {
                 
                // If row number is odd print
                // the row in reversed order
                cout << k << " ";
 
            }
            element_value += N;
        }
        cout << endl;
        i = i + 1;
    }
}
 
// Driver Code
int main()
{
     
    // Given matrix size
    int N = 4;
 
    // Function call
    UniqueMatrix(N);
}
 
// This code is contributed by chitranayal


Java




// Java program for the above approach
public class Gfg
{
   
    // Function to print the desired matrix
    public static void UniqueMatrix(int N)
    {
        int element_value = 1;
        int i = 0;
         
        // element_value will start from 1
        // and go up to N ^ 2
   
        // i is row number and it starts
        // from 0 and go up to N-1
       
        // Iterate ove all [0, N]
        while(i < N)
        {
            // If is even
            if(i % 2 == 0)
            {
                for(int f = element_value;
                    f < element_value + N; f++)
                {
                   
                    // If row number is even print
                    // the row in forward order
                    System.out.print(f+" ");
                }
                element_value += N;
            }
            else
            {
                for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                   
                    // If row number is odd print
                    // the row in reversed order
                    System.out.print(k+" ");
                }
                element_value += N;
            }
            System.out.println();
            i = i + 1;
        }
    }
   
    // Driver Code
    public static void main(String []args)
    {
       
        // Given matrix size
        int N = 4;
       
        // Function call
        UniqueMatrix(N);
    }
}
 
// This code is contributed by avanitrachhadiya2155


Python3




# Python3 program for the above approach
 
# Function to print the desired matrix
def UniqueMatrix(N):
 
    element_value = 1
    i = 0
 
    # element_value will start from 1
    # and go up to N ^ 2
 
    # i is row number and it starts
    # from 0 and go up to N-1
     
    # Iterate ove all [0, N]
    while(i < N):
         
        # If is even
        if(i % 2 == 0):
 
            for f in range(element_value, element_value + N, 1):
 
                # If row number is even print
                # the row in forward order
                print(f, end =' ')
            element_value += N
 
        else:
 
            for k in range(element_value + N-1, element_value-1, -1):
 
                # if row number is odd print
                # the row in reversed order
                print(k, end =' ')
 
            element_value += N
 
        print()
        i = i + 1
 
 
# Driver Code
 
# Given Matrix Size
N = 4
 
# Function Call
UniqueMatrix(N)


C#




// C# program for the above approach
using System;
class GFG
{
    static void UniqueMatrix(int N)
    {
        int element_value = 1;
        int i = 0;
          
        // element_value will start from 1
        // and go up to N ^ 2
    
        // i is row number and it starts
        // from 0 and go up to N-1
        
        // Iterate ove all [0, N]
        while(i < N)
        {
           
            // If is even
            if(i % 2 == 0)
            {
                for(int f = element_value;
                    f < element_value + N; f++)
                {
                   
                    // If row number is even print
                    // the row in forward order
                    Console.Write(f + " ");
                }
                element_value += N;
            }
            else
            {
                for(int k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                    Console.Write(k + " ");
                }
                element_value += N;
            }
            Console.WriteLine();
            i = i + 1;
        }
    }
   
  // Driver code
  static public void Main ()
  {
    // Given matrix size
    int N = 4;
 
    // Function call
    UniqueMatrix(N);
  }
}
 
// This code is contributed by rag2127


Javascript




<script>
// Java script  program for the above approach
 
    // Function to print the desired matrix
    function  UniqueMatrix( N)
    {
        let element_value = 1;
        let i = 0;
         
        // element_value will start from 1
        // and go up to N ^ 2
 
        // i is row number and it starts
        // from 0 and go up to N-1
     
        // Iterate ove all [0, N]
        while(i < N)
        {
            // If is even
            if(i % 2 == 0)
            {
                for(let f = element_value;
                    f < element_value + N; f++)
                {
                 
                    // If row number is even print
                    // the row in forward order
                    document.write(f+" ");
                }
                element_value += N;
            }
            else
            {
                for(let k = element_value + N - 1;
                    k > element_value - 1; k--)
                {
                 
                    // If row number is odd print
                    // the row in reversed order
                    document.write(k+" ");
                }
                element_value += N;
            }
            document.write("<br>");
            i = i + 1;
        }
    }
 
    // Driver Code
     
        // Given matrix size
        let N = 4;
     
        // Function call
        UniqueMatrix(N);
 
// contributed by sravan kumar
</script>


Output: 

1 2 3 4 
8 7 6 5 
9 10 11 12 
16 15 14 13

 

Time Complexity: O(N^2) 
Auxiliary Space: O(N^2) 
 




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!