# Find all distinct subsets of a given set using BitMasking Approach

Given a set of positive integers, find all its subsets. The set can contain duplicate elements, so any repeated subset should be considered only once in the output.**Examples:**

Input:S = {1, 2, 2}Output:{}, {1}, {2}, {1, 2}, {2, 2}, {1, 2, 2}Explanation:The total subsets of given set are - {}, {1}, {2}, {2}, {1, 2}, {1, 2}, {2, 2}, {1, 2, 2} Here {2} and {1, 2} are repeated twice so they are considered only once in the output

**Prerequisite: **Power Set

The idea is to use a bit-mask pattern to generate all the combinations as discussed in previous post. But previous post will print duplicate subsets if the elements are repeated in the given set. To handle duplicate elements, we construct a string out of given subset such that subsets having similar elements will result in same string. We maintain a list of such unique strings and finally we decode all such string to print its individual elements.

Below is its implementation –

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## C++

`// C++ program to find all subsets of given set. Any` `// repeated subset is considered only once in the output` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Utility function to split the string using a delim. Refer -` `vector<string> split(` `const` `string &s, ` `char` `delim)` `{` ` ` `vector<string> elems;` ` ` `stringstream ss(s);` ` ` `string item;` ` ` `while` `(getline(ss, item, delim))` ` ` `elems.push_back(item);` ` ` `return` `elems;` `}` `// Function to find all subsets of given set. Any repeated` `// subset is considered only once in the output` `int` `printPowerSet(` `int` `arr[], ` `int` `n)` `{` ` ` `vector<string> list;` ` ` `/* Run counter i from 000..0 to 111..1*/` ` ` `for` `(` `int` `i = 0; i < (` `int` `) ` `pow` `(2, n); i++)` ` ` `{` ` ` `string subset = ` `""` `;` ` ` `// consider each element in the set` ` ` `for` `(` `int` `j = 0; j < n; j++)` ` ` `{` ` ` `// Check if jth bit in the i is set. If the bit` ` ` `// is set, we consider jth element from set` ` ` `if` `((i & (1 << j)) != 0)` ` ` `subset += to_string(arr[j]) + ` `"|"` `;` ` ` `}` ` ` `// if subset is encountered for the first time` ` ` `// If we use set<string>, we can directly insert` ` ` `if` `(find(list.begin(), list.end(), subset) == list.end())` ` ` `list.push_back(subset);` ` ` `}` ` ` `// consider every subset` ` ` `for` `(string subset : list)` ` ` `{` ` ` `// split the subset and print its elements` ` ` `vector<string> arr = split(subset, ` `'|'` `);` ` ` `for` `(string str: arr)` ` ` `cout << str << ` `" "` `;` ` ` `cout << endl;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `arr[] = { 10, 12, 12 };` ` ` `int` `n = ` `sizeof` `(arr)/` `sizeof` `(arr[0]);` ` ` `printPowerSet(arr, n);` ` ` `return` `0;` `}` |

## Python3

`# Python3 program to find all subsets of` `# given set. Any repeated subset is` `# considered only once in the output` `def` `printPowerSet(arr, n):` ` ` ` ` `# Function to find all subsets of given set.` ` ` `# Any repeated subset is considered only` ` ` `# once in the output` ` ` `_list ` `=` `[]` ` ` `# Run counter i from 000..0 to 111..1` ` ` `for` `i ` `in` `range` `(` `2` `*` `*` `n):` ` ` `subset ` `=` `""` ` ` `# consider each element in the set` ` ` `for` `j ` `in` `range` `(n):` ` ` `# Check if jth bit in the i is set.` ` ` `# If the bit is set, we consider` ` ` `# jth element from set` ` ` `if` `(i & (` `1` `<< j)) !` `=` `0` `:` ` ` `subset ` `+` `=` `str` `(arr[j]) ` `+` `"|"` ` ` `# if subset is encountered for the first time` ` ` `# If we use set<string>, we can directly insert` ` ` `if` `subset ` `not` `in` `_list ` `and` `len` `(subset) > ` `0` `:` ` ` `_list.append(subset)` ` ` `# consider every subset` ` ` `for` `subset ` `in` `_list:` ` ` `# split the subset and print its elements` ` ` `arr ` `=` `subset.split(` `'|'` `)` ` ` `for` `string ` `in` `arr:` ` ` `print` `(string, end ` `=` `" "` `)` ` ` `print` `()` `# Driver Code` `if` `__name__ ` `=` `=` `'__main__'` `:` ` ` `arr ` `=` `[` `10` `, ` `12` `, ` `12` `]` ` ` `n ` `=` `len` `(arr)` ` ` `printPowerSet(arr, n)` `# This code is contributed by vibhu4agarwal` |

**Output:**

10 12 10 12 12 12 10 12 12

Time Complexity of this approach is: (N*2^{N}).

Refer to the below article to solve the problem using the backtracking approach.

https://www.geeksforgeeks.org/backtracking-to-find-all-subsets/

This article is contributed by **Aditya Goel**. 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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