# 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.

## 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 contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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