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

Last Updated : 18 Sep, 2023

Given 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 2^x - 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 2^u - 1 + 2^v - 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.

Implementation:

CPP

// 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 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 += 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++) {
            if (a[j][i] == 1)
                u++;
            else
                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;
}

Java

// Java program to compute number of sets
// in a binary matrix.
import java.util.*;
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.

Python3

# Python3 program to compute number of sets 
# in a binary matrix. 
m = 3 # no of columns 
n = 2 # no of rows 

# function to calculate the number of 
# non empty sets of cell 
def countSets(a):
    
    # stores the final answer 
    res = 0
    
    # traverses row-wise 
    for i in range(n):
        u = 0
        v = 0
        for j in range(m):
            if a[i][j]:
                u += 1
            else: 
                v += 1
        res += pow(2, u) - 1 + pow(2, v) - 1
    
    # traverses column wise 
    for i in range(m):
        
        u = 0
        v = 0
        for j in range(n):
            if a[j][i]:
                u += 1
            else: 
                v += 1
        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. 
a = [[1, 0, 1],[0, 1, 0]] 

print(countSets(a))

# This code is contributed by shubhamsingh10

// 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.

PHP

<?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.
?>

JavaScript

<script>
// javascript program to compute number of sets
// in a binary matrix.
     var m = 3; // no of columns
     var n = 2; // no of rows

    // function to calculate the number of
    // non empty sets of cell
    function countSets(a) {

        // stores the final answer
        var res = 0;

        // traverses row-wise
        for (i = 0; i < n; i++) {
            var u = 0, v = 0;
            for (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 (i = 0; i < m; i++) {
            var u = 0, v = 0;
            for (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
    
        var a = [ [ 1, 0, 1 ], [ 0, 1, 0 ] ];

        document.write(countSets(a));

// This code is contributed by Rajput-Ji
</script>

Output

Time Complexity: O(N*M), as we are using nested loops to traverse N*M times.
Auxiliary Space: O(1), as we are not using any extra space.

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