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Finding all subsets of a given set in Java

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  • Difficulty Level : Medium
  • Last Updated : 30 Jun, 2022

Problem: Find all the subsets of a given set.

Input: 
S = {a, b, c, d}
Output:
{}, {a} , {b}, {c}, {d}, {a,b}, {a,c},
{a,d}, {b,c}, {b,d}, {c,d}, {a,b,c}, 
{a,b,d}, {a,c,d}, {b,c,d}, {a,b,c,d}

The total number of subsets of any given set is equal to 2^ (no. of elements in the set). If we carefully notice it is nothing but binary numbers from 0 to 15 which can be shown as below: 
 

0000 
 
{} 
 
0001 
 
{a} 
 
0010 
 
{b} 
 
0011 
 
{a, b} 
 
0100 
 
{c} 
 
0101 
 
{a, c} 
 
0110 
 
{b, c} 
 
0111 
 
{a, b, c} 
 
1000 
 
{d} 
 
1001 
 
{a, d} 
 
1010 
 
{b, d} 
 
1011 
 
{a, b, d} 
 
1100 
 
{c, d} 
 
1101 
 
{a, c, d} 
 
1110 
 
{b, c, d} 
 
1111 
 
{a, b, c, d} 
 

 

Starting from right, 1 at ith position shows that the ith element of the set is present as 0 shows that the element is absent. Therefore, what we have to do is just generate the binary numbers from 0 to 2^n – 1, where n is the length of the set or the numbers of elements in the set.

Implementation:

Java




// A Java program to print all subsets of a set
import java.io.IOException;
 
class Main
{
    // Print all subsets of given set[]
    static void printSubsets(char set[])
    {
        int n = set.length;
 
        // Run a loop for printing all 2^n
        // subsets one by one
        for (int i = 0; i < (1<<n); i++)
        {
            System.out.print("{ ");
 
            // Print current subset
            for (int j = 0; j < n; j++)
 
                // (1<<j) is a number with jth bit 1
                // so when we 'and' them with the
                // subset number we get which numbers
                // are present in the subset and which
                // are not
                if ((i & (1 << j)) > 0)
                    System.out.print(set[j] + " ");
 
            System.out.println("}");
        }
    }
 
    // Driver code
    public static void main(String[] args)
    {
        char set[] = {'a', 'b', 'c'};
        printSubsets(set);
    }
}

Output

{ }
{ a }
{ b }
{ a b }
{ c }
{ a c }
{ b c }
{ a b c }

Time complexity: O(n * (2^n)) as the outer loop runs for O(2^n) and the inner loop runs for O(n).

Related Post: 
Finding all subsets of a Set in C/C++

This article is contributed by Nikhil Tekwani. If you like GeeksforGeeks and would like to contribute, you can also write an article and mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks. 


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