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

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 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Approach: To solve this problem lets 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 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

Output:

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

Time Complexity: O(N2)

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.

My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.