Recursive program to generate power set (original) (raw)

Last Updated : 11 Jul, 2025

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" }

**Using Pick and Do Not Pick for Every Element:

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 C++ `

// CPP program to generate power set #include <bits/stdc++.h> 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

Python

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

`

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

Fixing Prefixes One by One

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 <bits/stdc++.h> 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

Python

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

`

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

**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 newSubset
  • Add newSubset into moreSubsets
• Add moreSubsets into allSubsets
• 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 <bits/stdc++.h> using namespace std;

vector<vector > getSubset(vector set, int index) { vector<vector > allSubsets; if (index < 0) { vector v; allSubsets.push_back(v); }

else { allSubsets = getSubset(set, index - 1); string item = set[index]; vector<vector > moreSubsets;

for (vector<string> subset : allSubsets) {
  vector<string> 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<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<ArrayList<String> > result
        = getSubset(set, index);
    System.out.println(result);
}

static ArrayList<ArrayList<String> >
getSubset(String[] set, int index)
{
    ArrayList<ArrayList<String> > allSubsets;
    if (index < 0) {
        allSubsets
            = new ArrayList<ArrayList<String> >();
        allSubsets.add(new ArrayList<String>());
    }

    else {
        allSubsets = getSubset(set, index - 1);
        String item = set[index];
        ArrayList<ArrayList<String> > moreSubsets
            = new ArrayList<ArrayList<String> >();

        for (ArrayList<String> subset : allSubsets) {
            ArrayList<String> newSubset
                = new ArrayList<String>();
            newSubset.addAll(subset);
            newSubset.add(item);
            moreSubsets.add(newSubset);
        }
        allSubsets.addAll(moreSubsets);
    }
    return allSubsets;
}

}

Python

Python recursive code to print

all subsets of set using ArrayList

def get_subset(s, index): all_subsets = [] if index < 0: all_subsets.append([]) else: all_subsets = get_subset(s, index-1) item = s[index] more_subsets = [] for subset in all_subsets: new_subset = [] for i in subset: new_subset.append(i) new_subset.append(item) more_subsets.append(new_subset) for i in more_subsets: all_subsets.append(i) return all_subsets

Driver Code

set = ["a", "b", "c"] index = len(set) - 1 result = get_subset(set, index)

for subset in result: print("[", end=" ") for item in subset: print(item, end=" ") print("],", end=" ")

C#

// c# code for the above approach using System; using System.Collections.Generic;

namespace Subsets { public class GFG { static List<List > GetSubsets(List set, int index) { List<List > allSubsets; if (index < 0) { List emptyList = new List(); allSubsets = new List<List >{ emptyList }; } else { allSubsets = GetSubsets(set, index - 1); string item = set[index]; List<List > moreSubsets = new List<List >(); foreach(List subset in allSubsets) { List newSubset = new List(); foreach(string it in subset) { newSubset.Add(it); } newSubset.Add(item); moreSubsets.Add(newSubset); } foreach(List it in moreSubsets) { allSubsets.Add(it); } } return allSubsets; }

static void Main(string[] args)
{
    List<string> set
        = new List<string>{ "a", "b", "c" };
    int index = set.Count - 1;
    List<List<string> > result = GetSubsets(set, index);
    foreach(List<string> it in result)
    {
        Console.Write("[ ");
        foreach(string itr in it)
        {
            Console.Write(itr + " ");
        }
        Console.Write("], ");
    }
}

} }

JavaScript

//javascript code for the above approach function getSubset(set, index) { let allSubsets = [];

if (index < 0) { let v = []; allSubsets.push(v); } else { allSubsets = getSubset(set, index - 1); let item = set[index]; let moreSubsets = [];

for (let subset of allSubsets) {
  let newSubset = [];
  for (let it of subset)
    newSubset.push(it);
  newSubset.push(item);
  moreSubsets.push(newSubset);
}

for (let it of moreSubsets)
  allSubsets.push(it);

}

return allSubsets; }

let set = ["a", "b", "c"]; let index = set.length - 1; let result = getSubset(set, index);

for (let it of result) { console.log(" [ " + it.join(",") + " ],"); }

`

Output

[[], [a], [b], [a, b], [c][/c], [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.