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 matrixvoid 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 Codeint main(){ int N = 5; constructBinaryMat(N); return 0;} |
Java
// Java program to implement// the above approachclass GFG{// Function to construct// the required binary matrixstatic 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 Codepublic 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 matrixdef 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 Codeif __name__ == '__main__': N = 5 constructBinaryMat(N) # This code is contributed by mohit kumar 29 |
C#
// C# program to implement// the above approachusing System;class GFG{// Function to construct// the required binary matrixstatic 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 Codepublic 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 matrixfunction 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 Codevar 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)
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
