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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.



C++

// 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;     
}  

Java

// Java implementation to count subsets having
// even numbers only and all are distinct
import java.util.*;
  
class GFG 
{
  
// function to count the
// required subsets
static int countSubsets(int arr[], int n)
{
    HashSet<Integer> us = new HashSet<>();
    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.add(arr[i]);
      
      
    // counting distinct even numbers
    even_count=us.size();
      
    // total count of required subsets
    return (int) (Math.pow(2, even_count) - 1);
}
  
// Driver code
public static void main(String[] args) 
{
    int arr[] = {4, 2, 1, 9, 2, 6, 5, 3};
    int n = arr.length;
    System.out.println("Number of subsets = "
        + countSubsets(arr, n));
}
}
  
// This code contributed by Rajput-Ji

Python3

# python implementation to count subsets having 
# even numbers only and all are distinct 
  
#function to count the required subsets 
def countSubSets(arr, n):
    us = set()
    even_count = 0
  
    # inserting even numbers in the set 'us' 
    # single copy of each number is retained 
    for i in range(n):
        if arr[i] % 2 == 0:
            us.add(arr[i])
  
    # counting distinct even numbers 
    for i in us:
        even_count += 1
  
    # total count of required subsets 
    return pow(2, even_count)-  1
  
  
# Driver program
arr = [4, 2, 1, 9, 2, 6, 5, 3]
n = len(arr)
print("Numbers of subset=", countSubSets(arr,n))
  
# This code is contributed by Shrikant13
  


Output:

Number of subsets = 7

Time Complexity: O(n)

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



This article is attributed to GeeksforGeeks.org

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