Given a number N, the task is to create a square matrix of size N*N with values in range [1, N*N], such that the sum of each diagonal of an even sub-square matrix is even.
Examples:
Input: N = 3
Output:
1 2 3
4 5 6
7 8 9
Explanation:For each even sub-square matrix the sum of each diagonal is a even number.
1 2
4 5
sum of each diagonal is 6 and 6 i.e even number.Input: N = 4
Output:
1 2 3 4
6 5 8 7
9 10 11 12
14 13 16 15
Explanation:
For each even sub-square matrix the sum of each diagonal is a even number.
1 2
6 5
sum of each diagonal is 6 and 8 i.e even number.
Approach: The idea is to arrange elements from 1 to N*N in the below-given ways:
- Initialize odd and even by 1 and 2 elements respectively.
- Iterate two nested loop in the range [0, N].
- If the sum of indices in the two nested loops is even the print the value of odd and increment odd by 2 and if the sum is odd then print the value of even, and increment even by 2.
Below is the implementation of the above approach:
// C++ program for the above approach #include <iostream> using namespace std;
// Function to print N*N order matrix // with all sub-matrix of even order // is sum of its diagonal also even void evenSubMatrix( int N)
{ // Even index
int even = 1;
// Odd index
int odd = 2;
// Iterate two nested loop
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++) {
// For even index the element
// should be consecutive odd
if ((i + j) % 2 == 0) {
cout << even << " " ;
even += 2;
}
// for odd index the element
// should be consecutive even
else {
cout << odd << " " ;
odd += 2;
}
}
cout << "\n" ;
}
} // Driver Code int main()
{ // Given order of matrix
int N = 4;
// Function call
evenSubMatrix(N);
return 0;
} |
// Java program for the above approach import java.util.*;
class GFG{
// Function to print N*N order matrix // with all sub-matrix of even order // is sum of its diagonal also even static void evenSubMatrix( int N)
{ // Even index
int even = 1 ;
// Odd index
int odd = 2 ;
// Iterate two nested loop
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < N; j++)
{
// For even index the element
// should be consecutive odd
if ((i + j) % 2 == 0 )
{
System.out.print(even + " " );
even += 2 ;
}
// For odd index the element
// should be consecutive even
else
{
System.out.print(odd + " " );
odd += 2 ;
}
}
System.out.println();
}
} // Driver code public static void main(String[] args)
{ // Given order of matrix
int N = 4 ;
// Function call
evenSubMatrix(N);
} } // This code is contributed by offbeat |
# Python3 program for the above approach # Function to print N*N order matrix # with all sub-matrix of even order # is sum of its diagonal also even def evenSubMatrix(N):
# Even index
even = 1
# Odd index
odd = 2
# Iterate two nested loop
for i in range (N):
for j in range (N):
# For even index the element
# should be consecutive odd
if ((i + j) % 2 = = 0 ):
print (even, end = " " )
even + = 2
# For odd index the element
# should be consecutive even
else :
print (odd, end = " " )
odd + = 2
print ()
# Driver Code # Given order of matrix N = 4
# Function call evenSubMatrix(N) # This code is contributed by sanjoy_62 |
// C# program for the above approach using System;
class GFG{
// Function to print N*N order matrix // with all sub-matrix of even order // is sum of its diagonal also even static void evenSubMatrix( int N)
{ // Even index
int even = 1;
// Odd index
int odd = 2;
// Iterate two nested loop
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
// For even index the element
// should be consecutive odd
if ((i + j) % 2 == 0)
{
Console.Write(even + " " );
even += 2;
}
// For odd index the element
// should be consecutive even
else
{
Console.Write(odd + " " );
odd += 2;
}
}
Console.WriteLine();
}
} // Driver code public static void Main(String[] args)
{ // Given order of matrix
int N = 4;
// Function call
evenSubMatrix(N);
} } // This code is contributed by amal kumar choubey |
<script> // Java script program for the above approach // Function to print N*N order matrix // with all sub-matrix of even order // is sum of its diagonal also even function evenSubMatrix( N)
{ // Even index
let even = 1;
// Odd index
let odd = 2;
// Iterate two nested loop
for (let i = 0; i < N; i++)
{
for (let j = 0; j < N; j++)
{
// For even index the element
// should be consecutive odd
if ((i + j) % 2 == 0)
{
document.write(even + " " );
even += 2;
}
// For odd index the element
// should be consecutive even
else
{
document.write(odd + " " );
odd += 2;
}
}
document.write( "<br>" );
}
} // Driver code // Given order of matrix
let N = 4;
// Function call
evenSubMatrix(N);
// This code is contributed by manoj </script> |
1 2 3 4 6 5 8 7 9 10 11 12 14 13 16 15
Time Complexity: O(N*N)
Auxiliary Space: O(1)