Find a way to fill matrix with 1’s and 0’s in blank positions

Given an N * M matrix mat[][] which consists of two types of characters ‘.’ and ‘_’. The task is to fill the matrix in positions where it contains ‘.’ with 1‘s and 0‘s. Fill the matrix in such a way that no two adjacent cells contain the same number and print the modified matrix.

Examples:

Input: mat[][] = {{‘.’, ‘_’}, {‘_’, ‘.’}}
Output:
1 _
_ 1

Input: mat[][] = {{‘_’, ‘_’}, {‘_’, ‘_’}}
Output:
_ _
_ _
There is no place to fill the numbers.

Approach: An efficient approach is to fill the matrix in the following pattern:

10101010…
01010101…
10101010…

skipping ‘_’ characters whenever encountered.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
#define N 2
#define M 2
  
// Function to generate and
// print the required matrix
void Matrix(char a[N][M])
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++) {
  
            // Replace the '.'
            if (a[i][j] == '.')
                a[i][j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            cout << a[i][j] << " ";
        }
        cout << endl;
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
// Driver code
int main()
{
    char a[N][M] = { { '.', '_' },
                     { '_', '.' } };
  
    Matrix(a);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java implementation of the approach
import java.io.*;
  
class GFG 
{
  
static int N = 2;
static int M = 2;
  
// Function to generate and
// print the required matrix
static void Matrix(char a[][])
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
  
            // Replace the '.'
            if (a[i][j] == '.')
                a[i][j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            System.out.print( a[i][j] + " ");
        }
        System.out.println();
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
    // Driver code
    public static void main (String[] args) 
    {
        char a[][] = { { '.', '_' },
                    { '_', '.' } };
  
        Matrix(a);
    }
}
  
// This code is contributed by anuj_67..

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 implementation of the approach 
  
N = 2
M = 2
  
# Function to generate and 
# print the required matrix 
def Matrix(a) : 
    ch = '1'
  
    for i in range(N) :
        for j in range(M) :
  
            # Replace the '.' 
            if (a[i][j] == '.') :
                a[i][j] = ch; 
  
            # Toggle number 
            if (ch == '1') :
                ch == '0'
            else :
                ch = '1'
  
            print(a[i][j],end = " "); 
              
        print() 
  
        # For each row, change 
        # the starting number 
        if (i % 2 == 0) :
            ch = '0'
        else :
            ch = '1'
  
# Driver code 
if __name__ == "__main__"
  
    a = [
            [ '.', '_' ], 
            [ '_', '.' ],
        ]
  
    Matrix(a); 
      
# This code is contributed by AnkitRai01

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# implementation of the approach
using System;
      
class GFG 
{
  
static int N = 2;
static int M = 2;
  
// Function to generate and
// print the required matrix
static void Matrix(char [,]a)
{
    char ch = '1';
  
    for (int i = 0; i < N; i++) 
    {
        for (int j = 0; j < M; j++) 
        {
  
            // Replace the '.'
            if (a[i,j] == '.')
                a[i,j] = ch;
  
            // Toggle number
            ch = (ch == '1') ? '0' : '1';
  
            Console.Write( a[i,j] + " ");
        }
        Console.WriteLine();
  
        // For each row, change
        // the starting number
        if (i % 2 == 0)
            ch = '0';
        else
            ch = '1';
    }
}
  
// Driver code
public static void Main (String[] args) 
{
    char [,]a = { { '.', '_' },
                { '_', '.' } };
  
    Matrix(a);
}
}
  
// This code has been contributed by 29AjayKumar

chevron_right


Output:

1 _ 
_ 1


My Personal Notes arrow_drop_up

pawanasipugmailcom

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.



Improved By : vt_m, 29AjayKumar, AnkitRai01