Counting sets of 1s and 0s in a binary matrix

Given a n × m binary matrix, count the number of sets where a set can be formed one or more same values in a row or column.
Examples:

Input: 1 0 1
       0 1 0 
Output: 8 
Explanation: There are six one-element sets
(three 1s and three 0s). There are two two-
element sets, the first one consists of the
first and the third cells of the first row.
The second one consists of the first and the 
third cells of the second row. 

Input: 1 0
       1 1 
Output: 6

The number of non-empty subsets of x elements is 2x – 1. We traverse every row and calculate numbers of 1’s and 0’s cells. For every u zeros and v ones, total sets is 2u – 1 + 2v – 1. We then traverse all columns and compute same values and compute overall sum. We finally subtract m x n from the overall sum as single elements are considered twice.



CPP

filter_none

edit
close

play_arrow

link
brightness_4
code

// CPP program to compute number of sets
// in a binary matrix.
#include <bits/stdc++.h>
using namespace std;
  
const int m = 3; // no of columns
const int n = 2; // no of rows
  
// function to calculate the number of
// non empty sets of cell
long long countSets(int a[n][m])
{   
    // stores the final answer 
    long long res = 0;
      
    // traverses row-wise 
    for (int i = 0; i < n; i++)
    {
        int u = 0, v = 0; 
        for (int j = 0; j < m; j++) 
            a[i][j] ? u++ : v++;          
        res += pow(2,u)-1 + pow(2,v)-1; 
    }
      
    // traverses column wise 
    for (int i = 0; i < m; i++)
    {
        int u = 0, v = 0; 
        for (int j = 0; j < n; j++) 
             a[j][i] ? u++ : v++;  
        res += pow(2,u)-1 + pow(2,v)-1; 
    }
      
    // at the end subtract n*m as no of
    // single sets have been added twice.
    return res-(n*m);
}
  
// driver program to test the above function.
int main() {
      
    int a[][3] = {(1, 0, 1),
                  (0, 1, 0)};
      
    cout << countSets(a); 
      
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java program to compute number of sets
// in a binary matrix.
class GFG {
static final int m = 3; // no of columns
static final int n = 2; // no of rows
  
// function to calculate the number of
// non empty sets of cell
static long countSets(int a[][]) {
  
    // stores the final answer
    long res = 0;
  
    // traverses row-wise
    for (int i = 0; i < n; i++) {
    int u = 0, v = 0;
    for (int j = 0; j < m; j++) {
        if (a[i][j] == 1)
        u++;
        else
        v++;
    }
    res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
    }
  
    // traverses column wise
    for (int i = 0; i < m; i++) {
    int u = 0, v = 0;
    for (int j = 0; j < n; j++) {
        if (a[j][i] == 1)
        u++;
        else
        v++;
    }
    res += Math.pow(2, u) - 1 + Math.pow(2, v) - 1;
    }
  
    // at the end subtract n*m as no of
    // single sets have been added twice.
    return res - (n * m);
}
  
// Driver code
public static void main(String[] args) {
    int a[][] = {{1, 0, 1}, {0, 1, 0}};
  
    System.out.print(countSets(a));
}
}
// This code is contributed by Anant Agarwal.

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# program to compute number of
// sets in a binary matrix.
using System;
  
class GFG {
      
    static int m = 3; // no of columns
    static int n = 2; // no of rows
      
    // function to calculate the number of
    // non empty sets of cell
    static long countSets(int [,]a)
    {
      
        // stores the final answer
        long res = 0;
      
        // Traverses row-wise
        for (int i = 0; i < n; i++)
        {
            int u = 0, v = 0;
              
            for (int j = 0; j < m; j++)
            {
                if (a[i,j] == 1)
                    u++;
                else
                    v++;
            }
            res += (long)(Math.Pow(2, u) - 1
                       + Math.Pow(2, v)) - 1;
        }
      
        // Traverses column wise
        for (int i = 0; i < m; i++)
        {
            int u = 0, v = 0;
              
            for (int j = 0; j < n; j++)
            {
                if (a[j,i] == 1)
                    u++;
                else
                    v++;
            }
            res += (long)(Math.Pow(2, u) - 1
                       + Math.Pow(2, v)) - 1;
        }
      
        // at the end subtract n*m as no of
        // single sets have been added twice.
        return res - (n * m);
    }
      
    // Driver code
    public static void Main()
    {
        int [,]a = {{1, 0, 1}, {0, 1, 0}};
      
        Console.WriteLine(countSets(a));
    }
}
  
// This code is contributed by vt_m.

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php
// PHP program to compute
// number of sets
// in a binary matrix.
  
// no of columns
$m = 3; 
  
// no of rows
$n = 2; 
  
// function to calculate the number 
// of non empty sets of cell
function countSets($a)
    global $m, $n;
      
    // stores the final answer 
    $res = 0;
      
    // traverses row-wise 
    for ($i = 0; $i < $n; $i++)
    {
        $u = 0; $v = 0; 
        for ( $j = 0; $j < $m; $j++) 
            $a[$i][$j] ? $u++ : $v++;     
        $res += pow(2, $u) - 1 + pow(2, $v) - 1; 
    
      
    // traverses column wise 
    for ($i = 0; $i < $m; $i++)
    {
        $u = 0;$v = 0; 
        for ($j = 0; $j < $n; $j++) 
            $a[$j][$i] ? $u++ : $v++; 
        $res += pow(2, $u) - 1 + 
                pow(2, $v) - 1; 
    }
      
    // at the end subtract
    // n*m as no of single
    // sets have been added 
    // twice.
    return $res-($n*$m);
}
  
    // Driver Code
    $a = array(array(1, 0, 1),
               array(0, 1, 0));
      
    echo countSets($a); 
      
// This code is contributed by anuj_67.
?>

chevron_right



Output:

8

Time Complexity: O(n * m)

This article is contributed by Raj. 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 write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : vt_m



Article Tags :
Practice Tags :


Be the First to upvote.


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.