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 :
- The elements of the matrix M should be an integer between 1 and N^2.
- All elements of the matrix M are pairwise distinct.
- 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:
- 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 matrixvoid 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 Codeint main(){ // Given matrix size int N = 4; // Function call UniqueMatrix(N);}// This code is contributed by chitranayal |
Java
// Java program for the above approachpublic 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 matrixdef 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 SizeN = 4# Function CallUniqueMatrix(N) |
C#
// C# program for the above approachusing 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)
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