Finding all subsets of a given set in Java

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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.

// 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 obe
        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 }

 

Disclaimer: This does not belong to TechCodeBit, its an article taken from the below
source and credits.
source and credits: http://www.geeksforgeeks.org
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rakesh

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