Given an integer N, the task is to construct a binary matrix of size N * N such that the sum of each row and each column of the matrix is a prime number.
Examples:
Input: N = 2
Output:1 1 1 1Explanation:
Sum of 0th row = 1 + 1 = 2 (Prime number)
Sum of 1st row = 1 + 1 = 2 (Prime number)
Sum of 0th column = 1 + 1 = 2 (Prime number)
Sum of 1st column = 1 + 1 = 2 (Prime number)Input: N = 4
Output:1 0 0 1 0 1 1 0 0 1 1 0 1 0 0 1
Approach: Follow the steps below to solve the problem:
- Initialize a Binary matrix, say mat[][] of size N * N.
- Update all possible values of mat[i][i] to 1.
- Update all possible values of mat[i][N – i -1] to 1.
- If N is an odd number then update the value mat[N / 2][0] and mat[0][N / 2] to 1.
Below is the implementation of the above approach.
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> using namespace std;
// Function to construct // the required binary matrix void constructBinaryMat( int N)
{ // Stores binary value with row
// and column sum as prime number
int mat[N][N];
// initialize the binary matrix mat[][]
memset (mat, 0, sizeof (mat));
// Update all possible values of
// mat[i][i] to 1
for ( int i = 0; i < N; i++) {
mat[i][i] = 1;
}
// Update all possible values of
// mat[i][N - i -1]
for ( int i = 0; i < N; i++) {
mat[i][N - i - 1] = 1;
}
// Check if N is an odd number
if (N % 2 != 0) {
// Update mat[N / 2][0] to 1
mat[N / 2][0] = 1;
// Update mat[0][N / 2] to 1
mat[0][N / 2] = 1;
}
// Print required binary matrix
for ( int i = 0; i < N; i++) {
for ( int j = 0; j < N; j++) {
cout << mat[i][j] << " " ;
}
cout << endl;
}
} // Driver Code int main()
{ int N = 5;
constructBinaryMat(N);
return 0;
} |
Java
// Java program to implement // the above approach class GFG{
// Function to construct // the required binary matrix static void constructBinaryMat( int N)
{ // Stores binary value with row
// and column sum as prime number
int mat[][] = new int [N][N];
// Update all possible values
// of mat[i][i] to 1
for ( int i = 0 ; i < N; i++)
{
mat[i][i] = 1 ;
}
// Update all possible values
// of mat[i][N - i -1]
for ( int i = 0 ; i < N; i++)
{
mat[i][N - i - 1 ] = 1 ;
}
// Check if N is an odd
// number
if (N % 2 != 0 )
{
// Update mat[N / 2][0]
// to 1
mat[N / 2 ][ 0 ] = 1 ;
// Update mat[0][N / 2]
// to 1
mat[ 0 ][N / 2 ] = 1 ;
}
// Print required binary matrix
for ( int i = 0 ; i < N; i++)
{
for ( int j = 0 ; j < N; j++)
{
System.out.print(mat[i][j] + " " );
}
System.out.println();
}
} // Driver Code public static void main(String[] args)
{ int N = 5 ;
constructBinaryMat(N);
} } // This code is contributed by Chitranayal |
Python3
# Python3 program to implement # the above approach # Function to construct # the required binary matrix def constructBinaryMat(N):
# Stores binary value with row
# and column sum as prime number
mat = [[ 0 for i in range (N)]
for i in range (N)]
# Initialize the binary matrix mat[][]
# memset(mat, 0, sizeof(mat));
# Update all possible values of
# mat[i][i] to 1
for i in range (N):
mat[i][i] = 1
# Update all possible values of
# mat[i][N - i -1]
for i in range (N):
mat[i][N - i - 1 ] = 1
# Check if N is an odd number
if (N % 2 ! = 0 ):
# Update mat[N / 2][0] to 1
mat[N / / 2 ][ 0 ] = 1
# Update mat[0][N / 2] to 1
mat[ 0 ][N / / 2 ] = 1
# Print required binary matrix
for i in range (N):
for j in range (N):
print (mat[i][j], end = " " )
print ()
# Driver Code if __name__ = = '__main__' :
N = 5
constructBinaryMat(N)
# This code is contributed by mohit kumar 29 |
C#
// C# program to implement // the above approach using System;
class GFG{
// Function to construct // the required binary matrix static void constructBinaryMat( int N)
{ // Stores binary value with row
// and column sum as prime number
int [,]mat = new int [N, N];
// Update all possible values
// of mat[i,i] to 1
for ( int i = 0; i < N; i++)
{
mat[i, i] = 1;
}
// Update all possible values
// of mat[i,N - i -1]
for ( int i = 0; i < N; i++)
{
mat[i, N - i - 1] = 1;
}
// Check if N is an odd
// number
if (N % 2 != 0)
{
// Update mat[N / 2,0]
// to 1
mat[N / 2, 0] = 1;
// Update mat[0,N / 2]
// to 1
mat[0, N / 2] = 1;
}
// Print required binary matrix
for ( int i = 0; i < N; i++)
{
for ( int j = 0; j < N; j++)
{
Console.Write(mat[i, j] + " " );
}
Console.WriteLine();
}
} // Driver Code public static void Main(String[] args)
{ int N = 5;
constructBinaryMat(N);
} } // This code is contributed by gauravrajput1 |
Javascript
<script> // Javascript program to implement // the above approach // Function to construct // the required binary matrix function constructBinaryMat(N)
{ // Stores binary value with row
// and column sum as prime number
var mat = Array.from(Array(N), () =>
Array(N).fill(0));
// Update all possible values of
// mat[i][i] to 1
for ( var i = 0; i < N; i++)
{
mat[i][i] = 1;
}
// Update all possible values of
// mat[i][N - i -1]
for ( var i = 0; i < N; i++)
{
mat[i][N - i - 1] = 1;
}
// Check if N is an odd number
if (N % 2 != 0)
{
// Update mat[N / 2][0] to 1
mat[parseInt(N / 2)][0] = 1;
// Update mat[0][N / 2] to 1
mat[0][parseInt(N / 2)] = 1;
}
// Print required binary matrix
for ( var i = 0; i < N; i++)
{
for ( var j = 0; j < N; j++)
{
document.write( mat[i][j] + " " );
}
document.write( "<br>" );
}
} // Driver Code var N = 5;
constructBinaryMat(N); // This code is contributed by rutvik_56 </script> |
Output:
1 0 1 0 1 0 1 0 1 0 1 0 1 0 0 0 1 0 1 0 1 0 0 0 1
Time Complexity: O(N2)
Auxiliary Space: O(N2)