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++ 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 over 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 |
// Java program for the above approach public 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 over 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 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) |
// C# program for the above approach using 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 |
<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> |
1 2 3 4 8 7 6 5 9 10 11 12 16 15 14 13
Time Complexity: O(N^2)
Auxiliary Space: O(1)