# Recursive program to generate power set

Given a set represented as a string, write a recursive code to print all subsets of it. The subsets can be printed in any order.

Examples:

Input :  set = “abc”
Output : { “”, “a”, “b”, “c”, “ab”, “ac”, “bc”, “abc”}

Input : set = “abcd”
Output : { “”, “a” ,”ab” ,”abc” ,”abcd”, “abd” ,”ac” ,”acd”, “ad” ,”b”, “bc” ,”bcd” ,”bd” ,”c” ,”cd” ,”d” }

## Recursive program to generate power set using Backtracking:

The idea is to fix a prefix, and generate all subsets beginning with the current prefix. After all subsets with a prefix are generated, replace the last character with one of the remaining characters.

Follow the approach to implement the above idea:

• The base condition of the recursive approach is when the current index reaches the size of the given string (i.e, index == n), then return
• First, print the current subset
• Iterate over the given string from the current index (i.e, index) to less than the size of the string
• Appending the remaining characters to the current subset
• Make the recursive call for the next index.
• Once all subsets beginning with the initial “curr” are printed, remove the last character to consider a different prefix of subsets.

Follow the steps below to implement the above approach:

## C++

 `// CPP program to generate power set``#include ``using` `namespace` `std;` `// str : Stores input string``// curr : Stores current subset``// index : Index in current subset, curr``void` `powerSet(string str, ``int` `index = -1, string curr = ``""``)``{``    ``int` `n = str.length();` `    ``// base case``    ``if` `(index == n)``        ``return``;` `    ``// First print current subset``    ``cout << curr << ``"\n"``;` `    ``// Try appending remaining characters``    ``// to current subset``    ``for` `(``int` `i = index + 1; i < n; i++) {` `        ``curr += str[i];``        ``powerSet(str, i, curr);` `        ``// Once all subsets beginning with``        ``// initial "curr" are printed, remove``        ``// last character to consider a different``        ``// prefix of subsets.``        ``curr.erase(curr.size() - 1);``    ``}``    ``return``;``}` `// Driver code``int` `main()``{``    ``string str = ``"abc"``;``    ``powerSet(str);``    ``return` `0;``}`

## Java

 `// Java program to generate power set``import` `java.util.*;` `class` `GFG {` `    ``// str : Stores input string``    ``// curr : Stores current subset``    ``// index : Index in current subset, curr``    ``static` `void` `powerSet(String str, ``int` `index, String curr)``    ``{``        ``int` `n = str.length();` `        ``// base case``        ``if` `(index == n) {``            ``return``;``        ``}` `        ``// First print current subset``        ``System.out.println(curr);` `        ``// Try appending remaining characters``        ``// to current subset``        ``for` `(``int` `i = index + ``1``; i < n; i++) {``            ``curr += str.charAt(i);``            ``powerSet(str, i, curr);` `            ``// Once all subsets beginning with``            ``// initial "curr" are printed, remove``            ``// last character to consider a different``            ``// prefix of subsets.``            ``curr = curr.substring(``0``, curr.length() - ``1``);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"abc"``;``        ``int` `index = -``1``;``        ``String curr = ``""``;``        ``powerSet(str, index, curr);``    ``}``}` `// This code is contributed by PrinciRaj1992`

## Python3

 `# Python3 program to generate power set` `# str : Stores input string``# curr : Stores current subset``# index : Index in current subset, curr`  `def` `powerSet(str1, index, curr):``    ``n ``=` `len``(str1)` `    ``# base case``    ``if` `(index ``=``=` `n):``        ``return` `    ``# First print current subset``    ``print``(curr)` `    ``# Try appending remaining characters``    ``# to current subset``    ``for` `i ``in` `range``(index ``+` `1``, n):``        ``curr ``+``=` `str1[i]``        ``powerSet(str1, i, curr)` `        ``# Once all subsets beginning with``        ``# initial "curr" are printed, remove``        ``# last character to consider a different``        ``# prefix of subsets.``        ``curr ``=` `curr.replace(curr[``len``(curr) ``-` `1``], "")` `    ``return`  `# Driver code``if` `__name__ ``=``=` `'__main__'``:``    ``str` `=` `"abc"``    ``powerSet(``str``, ``-``1``, "")` `# This code is contributed by``# Surendra_Gangwar`

## C#

 `// C# program to generate power set``using` `System;` `class` `GFG {` `    ``// str : Stores input string``    ``// curr : Stores current subset``    ``// index : Index in current subset, curr``    ``static` `void` `powerSet(``string` `str, ``int` `index, ``string` `curr)``    ``{``        ``int` `n = str.Length;` `        ``// base case``        ``if` `(index == n) {``            ``return``;``        ``}` `        ``// First print current subset``        ``Console.WriteLine(curr);` `        ``// Try appending remaining characters``        ``// to current subset``        ``for` `(``int` `i = index + 1; i < n; i++) {``            ``curr += str[i];``            ``powerSet(str, i, curr);` `            ``// Once all subsets beginning with``            ``// initial "curr" are printed, remove``            ``// last character to consider a different``            ``// prefix of subsets.``            ``curr = curr.Substring(0, curr.Length - 1);``        ``}``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``string` `str = ``"abc"``;``        ``int` `index = -1;``        ``string` `curr = ``""``;``        ``powerSet(str, index, curr);``    ``}``}` `// This code is contributed by Ita_c.`

## Javascript

 ``

Output

```a
ab
abc
ac
b
bc
c```

Time Complexity: O(2n)
Auxiliary Space: O(n), For recursive call stack

## Recursive program to generate power set using Recursion:

The idea is to consider two cases for every character.
(i) Consider current character as part of current subset
(ii) Do not consider current character as part of the current subset.

Follow the approach to implement the above idea:

• The base condition of the recursive approach is when the current index reaches the size of the given string (i.e, index == n). then print the current string say, curr.
• Make two recursive calls for the two cases for every character
• We consider the character as part of the current subset
• We do not consider current character as part of the current subset

Follow the steps below to implement the above approach:

## C++

 `// CPP program to generate power set``#include ``using` `namespace` `std;` `// str : Stores input string``// curr : Stores current subset``// index : Index in current subset, curr``void` `powerSet(string str, ``int` `index = 0, string curr = ``""``)``{``    ``int` `n = str.length();` `    ``// base case``    ``if` `(index == n) {``        ``cout << curr << endl;``        ``return``;``    ``}` `    ``// Two cases for every character``    ``// (i) We consider the character``    ``// as part of current subset``    ``// (ii) We do not consider current``    ``// character as part of current``    ``// subset``    ``powerSet(str, index + 1, curr + str[index]);``    ``powerSet(str, index + 1, curr);``}` `// Driver code``int` `main()``{``    ``string str = ``"abc"``;``    ``powerSet(str);``    ``return` `0;``}`

## Java

 `// Java program to generate power set``class` `GFG {` `    ``// str : Stores input string``    ``// curr : Stores current subset``    ``// index : Index in current subset, curr``    ``static` `void` `powerSet(String str, ``int` `index, String curr)` `    ``{``        ``int` `n = str.length();` `        ``// base case``        ``if` `(index == n) {``            ``System.out.println(curr);``            ``return``;``        ``}` `        ``// Two cases for every character``        ``// (i) We consider the character``        ``// as part of current subset``        ``// (ii) We do not consider current``        ``// character as part of current``        ``// subset``        ``powerSet(str, index + ``1``, curr + str.charAt(index));``        ``powerSet(str, index + ``1``, curr);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``String str = ``"abc"``;``        ``int` `index = ``0``;``        ``String curr = ``""``;``        ``powerSet(str, index, curr);``    ``}``}``// This code is contributed by 29AjayKumar`

## Python3

 `# Python3 program to generate power set``def` `powerSet(string, index, curr):` `    ``# string : Stores input string``    ``# curr : Stores current subset``    ``# index : Index in current subset, curr``    ``if` `index ``=``=` `len``(string):``        ``print``(curr)``        ``return` `    ``powerSet(string, index ``+` `1``,``             ``curr ``+` `string[index])``    ``powerSet(string, index ``+` `1``, curr)`  `# Driver Code``if` `__name__ ``=``=` `"__main__"``:` `    ``s1 ``=` `"abc"``    ``index ``=` `0``    ``curr ``=` `""``    ``powerSet(s1, index, curr)` `# This code is contributed by Ekta Singh`

## C#

 `// C# program to generate power set``using` `System;` `class` `GFG {` `    ``// str : Stores input string``    ``// curr : Stores current subset``    ``// index : Index in current subset, curr``    ``static` `void` `powerSet(String str, ``int` `index, String curr)` `    ``{``        ``int` `n = str.Length;` `        ``// base case``        ``if` `(index == n) {``            ``Console.WriteLine(curr);``            ``return``;``        ``}` `        ``// Two cases for every character``        ``// (i) We consider the character``        ``// as part of current subset``        ``// (ii) We do not consider current``        ``// character as part of current``        ``// subset``        ``powerSet(str, index + 1, curr + str[index]);``        ``powerSet(str, index + 1, curr);``    ``}` `    ``// Driver code``    ``public` `static` `void` `Main()``    ``{``        ``String str = ``"abc"``;``        ``int` `index = 0;``        ``String curr = ``""``;``        ``powerSet(str, index, curr);``    ``}``}` `// This code is contributed by Rajput-Ji`

## Javascript

 ``

Output

```abc
ab
ac
a
bc
b
c```

Time Complexity: O(2n)
Auxiliary Space: O(n), For recursive call stack

## Recursive program to generate power set using Backtracking using Bottom Up Approach:

The idea is to pick each element one by one from the input set, then generate a subset for the same, and we follow this process recursively.

Follow the steps below to implement the above idea:

• The base condition for the recursive call, when the current index becomes negative then add empty list into the allSubsets.
• Make recursive call for the current index – 1, then we’ll receive allSubsets list from 0 to index -1
• Create a list of list moreSubsets
• Iterate over all the subsets that are received from above recursive call
• Copy all the subset into newSubset
• Add the current item into newSebset
• Finally, return allSubsets.

Follow the steps below to implement the above approach:

## C++

 `// C++ Recursive code to print``// all subsets of set using ArrayList``#include ``using` `namespace` `std;` `vector > getSubset(vector set,``                                  ``int` `index)``{``  ``vector > allSubsets;``  ``if` `(index < 0) {``    ``vector v;``    ``allSubsets.push_back(v);``  ``}` `  ``else` `{``    ``allSubsets = getSubset(set, index - 1);``    ``string item = set[index];``    ``vector > moreSubsets;` `    ``for` `(vector subset : allSubsets) {``      ``vector newSubset;``      ``for` `(``auto` `it : subset)``        ``newSubset.push_back(it);``      ``newSubset.push_back(item);``      ``moreSubsets.push_back(newSubset);``    ``}``    ``for` `(``auto` `it : moreSubsets)``      ``allSubsets.push_back(it);``  ``}``  ``return` `allSubsets;``}``int` `main()``{` `  ``vector set = { ``"a"``, ``"b"``, ``"c"` `};` `  ``int` `index = set.size() - 1;``  ``vector > result = getSubset(set, index);``  ``for` `(``auto` `it : result) {``    ``cout << ``" [ "``;``    ``for` `(``auto` `itr : it) {``      ``cout << itr << ``","``;``    ``}``    ``cout << ``"],"``;``  ``}``}` `// This code is contributed by garg28harsh.`

## Java

 `// Java Recursive code to print``// all subsets of set using ArrayList``import` `java.util.ArrayList;` `public` `class` `PowerSet {` `    ``public` `static` `void` `main(String[] args)``    ``{` `        ``String[] set = { ``"a"``, ``"b"``, ``"c"` `};` `        ``int` `index = set.length - ``1``;``        ``ArrayList > result``            ``= getSubset(set, index);``        ``System.out.println(result);``    ``}` `    ``static` `ArrayList >``    ``getSubset(String[] set, ``int` `index)``    ``{``        ``ArrayList > allSubsets;``        ``if` `(index < ``0``) {``            ``allSubsets``                ``= ``new` `ArrayList >();``            ``allSubsets.add(``new` `ArrayList());``        ``}` `        ``else` `{``            ``allSubsets = getSubset(set, index - ``1``);``            ``String item = set[index];``            ``ArrayList > moreSubsets``                ``= ``new` `ArrayList >();` `            ``for` `(ArrayList subset : allSubsets) {``                ``ArrayList newSubset``                    ``= ``new` `ArrayList();``                ``newSubset.addAll(subset);``                ``newSubset.add(item);``                ``moreSubsets.add(newSubset);``            ``}``            ``allSubsets.addAll(moreSubsets);``        ``}``        ``return` `allSubsets;``    ``}``}`

Output

`[[], [a], [b], [a, b], , [a, c], [b, c], [a, b, c]]`

Time Complexity: O(n*2n)
Auxiliary Space: O(n), For recursive call stack

Iterative program for the power set.

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