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; // distinct even numbers even_count = us.size(); // 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 even_count = len(us) # 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 |
C#
// C# implementation to count subsets having // even numbers only and all are distinct using System; using System.Collections.Generic; class GFG { // function to count the // required subsets static int countSubsets(int []arr, int n) { HashSet<int> us = new HashSet<int>(); 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.Count; // 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; Console.WriteLine("Number of subsets = " + countSubsets(arr, n)); } } // This code contributed by Rajput-Ji |
Output:
Number of subsets = 7
Time Complexity: O(n)
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