Construct a Matrix whose each row and column contains N and M 1s respectively

Construct a matrix whose each row contains N 1’s and each column contains M 1’s.

Examples:

Input: N = 3, M = 4
Output:
1 1 1
1 1 1
1 1 1
1 1 1 
Explanation: Each row contains N’s 1 and each column contains M’s 1.

Input: N = 0, M = 0
Output: 0

Approach: 

Follow the below steps to solve the problem:

• If N and M are 0 then simply print 0.
• If exactly one of N and M is 0, then it is impossible to make the grid.
• Else make the 2D vector or array of dimension MXN.
• Print the vector/array using nested loops.

Below is the implementation of the above approach.

C++

// C++ code to implement the approach

#include <bits/stdc++.h>
using namespace std;

// Function to construct grid with N and M
void constructGrid(int N, int M)
{
    if (N == 0 && M == 0) {
        cout << 0 << endl;
    }

    if (N == 0 || M == 0) {
        cout << "Grid not possible." << endl;
    }

    // Construct the grid
    vector<vector<int> > grid(M, vector<int>(N, 1));

    // Print the grid
    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++) {
            cout << grid[i][j] << " ";
        }
        cout << endl;
    }
}

// Driver Code
int main()
{
    // Testcase1
    int N = 5, M = 4;

    // Function call
    cout << N << " " << M << endl;
    constructGrid(N, M);
    cout << endl;

    // Testcase2
    N = 3, M = 7;
    cout << N << " " << M << endl;
    constructGrid(N, M);

    return 0;
}

Java

// Java code to implement the approach

import java.io.*;

class GFG {

    // Function to construct grid with N and M
    static void constructGrid(int N, int M)
    {
        if (N == 0 && M == 0) {
            System.out.println(0);
        }

        if (N == 0 || M == 0) {
            System.out.println("Grid not possible");
        }

        // Construct the grid
        int[][] grid = new int[M][N];

        // Print the grid
        for (int i = 0; i < M; i++) {
            for (int j = 0; j < N; j++) {
                grid[i][j] = 1;
                System.out.print(grid[i][j] + " ");
            }
            System.out.println();
        }
    }

    public static void main(String[] args)
    {
        // Testcase1
        int N = 5, M = 4;

        // Function call
        System.out.println(N + " " + M);
        constructGrid(N, M);
        System.out.println();

        // Testcase2
        N = 3;
        M = 7;

        // Function call
        System.out.println(N + " " + M);
        constructGrid(N, M);
    }
}

// This code is contributed by lokesh

Python3

# Python code to implement the approach

# Function to construct grid with N and M
def constructGrid(N, M):
    if (N == 0 and M == 0):
        print(0)

    if (N == 0 or M == 0):
        print("Grid not possible.")

    # Construct the grid
    grid = [[1 for i in range(N)] for j in range(M)]

    # Print the grid
    for i in range(M):
        for j in range(N):
            print(grid[i][j], end=" ")
        print()


# Driver Code
if __name__ == '__main__':

    # Testcase1
    N = 5
    M = 4

    # Function call
    print(N, M)
    constructGrid(N, M)
    print()

    # Testcase2
    N = 3
    M = 7
    print(N, M)
    constructGrid(N, M)

# This code is contributed by Tapesh(tapeshdua420)

C#

// C# code to implement the approach
using System;

public class GFG {

  // Function to construct grid with N and M
  static void constructGrid(int N, int M)
  {
    if (N == 0 && M == 0) {
      Console.WriteLine(0);
    }

    if (N == 0 || M == 0) {
      Console.WriteLine("Grid not possible");
    }

    // Construct the grid
    int[, ] grid = new int[M, N];

    // Print the grid
    for (int i = 0; i < M; i++) {
      for (int j = 0; j < N; j++) {
        grid[i, j] = 1;
        Console.Write(grid[i, j] + " ");
      }
      Console.WriteLine();
    }
  }

  static public void Main()
  {

    // Code
    // Testcase1
    int N = 5, M = 4;

    // Function call
    Console.WriteLine(N + " " + M);
    constructGrid(N, M);
    Console.WriteLine();

    // Testcase2
    N = 3;
    M = 7;

    // Function call
    Console.WriteLine(N + " " + M);
    constructGrid(N, M);
  }
}

// This code is contributed by lokeshmvs21.

Javascript

// JavaScript code to implement the approach 

// Function to construct grid with N and M 
function constructGrid(N, M)  { 
    if (N == 0 && M == 0) { 
        console.log(0); 
    } 
    
    if (N == 0 || M == 0) { 
        console.log("Grid not possible."); 
    } 
    
    // Construct the grid 
    var grid = new Array(M); 
    for (var i = 0; i < M; i++) { 
        grid[i] = new Array(N); 
    } 
    
    for (var i = 0; i < M; i++) { 
        for (var j = 0; j < N; j++) { 
            grid[i][j] = 1; 
        } 
    } 
    
    // Print the grid 
    for (var i = 0; i < M; i++) { 
        for (var j = 0; j < N; j++) { 
            process.stdout.write(grid[i][j]+" "); 
        } 
        console.log(); 
    } 
} 

// Driver Code 

// Testcase1 
var N = 5; 
var M = 4; 

// Function call 
console.log(N, M); 
constructGrid(N, M); 
console.log(); 

// Testcase2 
N = 3; 
M = 7; 
console.log(N, M); 
constructGrid(N, M); 

// This code is contributed by Tapesh(tapeshdua420)

Output

5 4
1 1 1 1 1 
1 1 1 1 1 
1 1 1 1 1 
1 1 1 1 1 

3 7
1 1 1 
1 1 1 
1 1 1 
1 1 1 
1 1 1 
1 1 1 
1 1 1 

Time Complexity: O(N*M)
Auxiliary Space: O(N*M)

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