Count subsets having distinct even numbers

Given a sequence of n numbers. The task is to count all the subsets of the given set which only have even numbers and all are distinct.
Note: By the property of sets, if two subsets have the same set of elements then they are considered as one. For example: [2, 4, 8] and [4, 2, 8] are considered to be the same.

Examples:

Input : {4, 2, 1, 9, 2, 6, 5, 3} 
Output : 7
The subsets are:
[4], [2], [6], [4, 2], 
[2, 6], [4, 6], [4, 2, 6]

Input : {10, 3, 4, 2, 4, 20, 10, 6, 8, 14, 2, 6, 9}
Output : 127

A simple approach is to consider all the subsets and check whether they satisfy the given conditions or not. The time complexity will be in exponential.

An efficient approach is to count number of distinct even numbers. Let this be ceven. And then apply formula:

2ceven – 1

This is similar to counting the number of subsets of a given set of n elements. 1 is subtracted because the null set is not considered.

[sourcecode language=”CPP”]
// C++ implementation to count subsets having
// even numbers only and all are distinct
#include <bits/stdc++.h>
using namespace std;

// function to count the
// required subsets
int countSubsets(int arr[], int n)
{
unordered_set<int> us;
int even_count = 0;

// inserting even numbers in the set ‘us’
// single copy of each number is retained
for (int i=0; i<n; i++)
if (arr[i] % 2 == 0)
us.insert(arr[i]);

unordered_set<int>:: iterator itr;

// counting distinct even numbers
for (itr=us.begin(); itr!=us.end(); itr++)
even_count++;

// total count of required subsets
return (pow(2, even_count) – 1);
}

// Driver program to test above
int main()
{
int arr[] = {4, 2, 1, 9, 2, 6, 5, 3};
int n = sizeof(arr) / sizeof(arr[0]);
cout << "Number of subsets = "
<< countSubsets(arr, n);
return 0;
}
[/sourcecode]
Output:

Number of subsets = 7

Time Complexity: O(n)

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