Finding all subsets of a given set in Java

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).

Auxiliary Space :O(1),  since no extra space is used.

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.