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:**2 ^{ceven} – 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)

This article is contributed by **Ayush Jauhari**. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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