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
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 <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 |
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
Time Complexity: O(N2)
Recommended Posts:
- Check if a grid can become row-wise and column-wise sorted after adjacent swaps
- Filling diagonal to make the sum of every row, column and diagonal equal of 3x3 matrix
- Find the number of squares inside the given square grid
- Find the number of p-sided squares in a grid with K blacks painted
- Program to find the Sum of each Row and each Column of a Matrix
- Find column with maximum sum in a Matrix
- Find maximum element of each column in a matrix
- Find a Square Matrix such that sum of elements in every row and column is K
- Given a Boolean Matrix, find k such that all elements in k'th row are 0 and k'th column are 1.
- Find if a binary matrix exists with given row and column sums
- Find probability of selecting element from kth column after N iterations
- Find the original matrix when largest element in a row and a column are given
- Find pair with maximum difference in any column of a Matrix
- Find sum of all elements in a matrix except the elements in row and/or column of given cell?
- Find Multiples of 2 or 3 or 5 less than or equal to N
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.