C++ Program for 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 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.

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 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;
}

Output: 

8

Time Complexity: O(n * m) 

Space Complexity: O(1) as no extra space has been taken.

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