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

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:

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: 
 

Below is the implementation of the above approach:
 

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// 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
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# 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)
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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) 
 

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Improved By : chitranayal, poulami21ghosh

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