Find an N x N grid whose xor of every row and column is equal

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Given an integer N which is a multiple of 4, the task is to find an N x N grid for which the bitwise xor of every row and column is the same.
Examples:

Input: N = 4
Output:
0 1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
Input: N = 8
Output:
0 1 2 3 16 17 18 19
4 5 6 7 20 21 22 23
8 9 10 11 24 25 26 27
12 13 14 15 28 29 30 31
32 33 34 35 48 49 50 51
36 37 38 39 52 53 54 55
40 41 42 43 56 57 58 59
44 45 46 47 60 61 62 63

Approach: To solve this problem let’s fix the xor of every row and column to 0 since xor of 4 consecutive numbers starting from 0 is 0. Here is an example of a 4 x 4 matrix:

0 ^ 1 ^ 2 ^ 3 = 0
4 ^ 5 ^ 6 ^ 7 = 0
8 ^ 9 ^ 10 ^ 11 = 0
12 ^ 13 ^ 14 ^ 15 = 0
and so on.

If you notice in the above example, the xor of every row and column is 0. Now we need to place the numbers in such a way that the xor of each row and column is 0. So we can divide our N x N matrix into smaller 4 x 4 matrices with N / 4 rows and columns and fill the cells in a way that the xor of every row and column is 0.
Below is the implementation of the above approach:

C++

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to find the n x n matrix
// that satisfies the given condition
void findGrid(int n)
{
    int arr[n][n];
 
    // Initialize x to 0
    int x = 0;
 
    // Divide the n x n matrix into n / 4 matrices
    // for each of the n / 4 rows where
    // each matrix is of size 4 x 4
    for (int i = 0; i < n / 4; i++) {
        for (int j = 0; j < n / 4; j++) {
            for (int k = 0; k < 4; k++) {
                for (int l = 0; l < 4; l++) {
                    arr[i * 4 + k][j * 4 + l] = x;
                    x++;
                }
            }
        }
    }
 
    // Print the generated matrix
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            cout << arr[i][j] << " ";
        }
        cout << "\n";
    }
}
 
// Driver code
int main()
{
    int n = 4;
 
    findGrid(n);
 
    return 0;
}

Java

// Java implementation of the approach
import java.io.*;
 
class GFG 
{
     
// Function to find the n x n matrix
// that satisfies the given condition
static void findGrid(int n)
{
    int [][]arr = new int[n][n];
 
    // Initialize x to 0
    int x = 0;
 
    // Divide the n x n matrix into n / 4 matrices
    // for each of the n / 4 rows where
    // each matrix is of size 4 x 4
    for (int i = 0; i < n / 4; i++)
    {
        for (int j = 0; j < n / 4; j++) 
        {
            for (int k = 0; k < 4; k++) 
            {
                for (int l = 0; l < 4; l++) 
                {
                    arr[i * 4 + k][j * 4 + l] = x;
                    x++;
                }
            }
        }
    }
 
    // Print the generated matrix
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            System.out.print(arr[i][j] + " ");
        }
        System.out.println(" ");
    }
}
 
// Driver code
public static void main (String[] args)
{
    int n = 4;
     
    findGrid(n);
}
}
 
// This code is contributed by ajit.

Python3

# Python3 implementation of the approach 
 
# Function to find the n x n matrix 
# that satisfies the given condition 
def findGrid(n): 
 
    arr = [[0 for k in range(n)] 
              for l in range(n)] 
 
    # Initialize x to 0 
    x = 0
 
    # Divide the n x n matrix into n / 4 matrices 
    # for each of the n / 4 rows where 
    # each matrix is of size 4 x 4 
    for i in range(n // 4): 
        for j in range(n // 4): 
            for k in range(4): 
                for l in range(4): 
                    arr[i * 4 + k][j * 4 + l] = x 
                    x += 1
 
    # Print the generated matrix 
    for i in range(n): 
        for j in range(n): 
            print(arr[i][j], end = " ")
        print()
 
# Driver code 
n = 4
findGrid(n) 
 
# This code is contributed by divyamohan123

C#

// C# implementation of the approach
using System;
     
class GFG 
{
     
// Function to find the n x n matrix
// that satisfies the given condition
static void findGrid(int n)
{
    int [,]arr = new int[n, n];
 
    // Initialize x to 0
    int x = 0;
 
    // Divide the n x n matrix into n / 4 matrices
    // for each of the n / 4 rows where
    // each matrix is of size 4 x 4
    for (int i = 0; i < n / 4; i++)
    {
        for (int j = 0; j < n / 4; j++) 
        {
            for (int k = 0; k < 4; k++) 
            {
                for (int l = 0; l < 4; l++) 
                {
                    arr[i * 4 + k, j * 4 + l] = x;
                    x++;
                }
            }
        }
    }
 
    // Print the generated matrix
    for (int i = 0; i < n; i++) 
    {
        for (int j = 0; j < n; j++) 
        {
            Console.Write(arr[i, j] + " ");
        }
        Console.WriteLine(" ");
    }
}
 
// Driver code
public static void Main (String[] args)
{
    int n = 4;
     
    findGrid(n);
}
}
 
// This code is contributed by PrinciRaj1992

Javascript

<script>
 
// Javascript implementation of the approach
 
// Function to find the n x n matrix
// that satisfies the given condition
function findGrid(n)
{
    let arr = new Array(n);
    for (let i = 0; i < n; i++)
        arr[i] = new Array(n);
 
    // Initialize x to 0
    let x = 0;
 
    // Divide the n x n matrix into n / 4 matrices
    // for each of the n / 4 rows where
    // each matrix is of size 4 x 4
    for (let i = 0; i < parseInt(n / 4); i++) {
        for (let j = 0; j < parseInt(n / 4); j++) {
            for (let k = 0; k < 4; k++) {
                for (let l = 0; l < 4; l++) {
                    arr[i * 4 + k][j * 4 + l] = x;
                    x++;
                }
            }
        }
    }
 
    // Print the generated matrix
    for (let i = 0; i < n; i++) {
        for (let j = 0; j < n; j++) {
            document.write(arr[i][j] + " ");
        }
        document.write("<br>");
    }
}
 
// Driver code
    let n = 4;
 
    findGrid(n);
 
</script>

Output:

0 1 2 3 
4 5 6 7 
8 9 10 11 
12 13 14 15

Time Complexity: O(N²)

Auxiliary Space: O(N²)

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