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Find all distinct subsets of a given set using BitMasking Approach

  • Difficulty Level : Hard
  • Last Updated : 07 Feb, 2022

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.

Note: This method will only work on sorted array. 
Below is its implementation –  

C++14




// 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;
 
// Function to find all subsets of given set. Any repeated
// subset is considered only once in the output
vector<vector<int> > findPowerSet(vector<int>& nums)
{
    int bits = nums.size();     // size of array to set bit
    unsigned int pow_set_size = pow(2, bits);     // total number of subsets = pow(2, sizeof(arr))
    sort(nums.begin(), nums.end());     // sort to avoid adding permutation of the substring
    vector<vector<int>> ans;
    vector<string> list;     // to store subset as a list to avoid adding exact duplicates
 
     
    // counter 000..0 to 111..1
    for (int counter = 0; counter < pow_set_size; counter++) {
        vector<int> subset;
        string temp = "";
     
        // check for the current bit in the counter
        for (int j = 0; j < bits; j++) {
            if (counter & (1 << j)) {
                subset.push_back(nums[j]);
                // add special character to separate integers
                temp += to_string(nums[j]) + '$';
            }
        }
     
        if (find(list.begin(), list.end(), temp) == list.end()) {
            ans.push_back(subset);
            list.push_back(temp);
        }
    }
 
    return ans;
}
 
// Driver code
int main()
{
    vector<int> arr{ 10,12,12 };
 
    vector<vector<int> > power_set = findPowerSet(arr);
     
    for(int i=0;i<power_set.size();i++)
    {
        for(int j=0;j<power_set[i].size();j++)
        cout<<power_set[i][j]<<" ";
        cout<<endl;
    }
 
    return 0;
}
// this code is contributed by prophet1999

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: O(N*2N)

Auxiliary Space: O(N*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.
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.
 


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