Recursive program to generate power set

Difficulty Level :
Medium
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" 
Method 1: The idea is to fix a prefix, 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.  
C++
Java
Python3
C#
Javascript
// 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, currvoid 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 codeint main(){    string str = "abc";    powerSet(str);    return 0;}
// Java program to generate power setimport 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 program to generate power set # str : Stores input string# curr : Stores current subset# index : Index in current subset, currdef 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 codeif __name__ == '__main__':    str = "abc";    powerSet(str, -1, "") # This code is contributed by# Surendra_Gangwar
// C# program to generate power setusing 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.
<script>// Javascript program to generate power set         // str : Stores input string    // curr : Stores current subset    // index : Index in current subset, curr    function powerSet(str,index,curr)    {        let n = str.length;           // base case        if (index == n)        {            return;        }           // First print current subset        document.write(curr+"<br>");           // Try appending remaining characters        // to current subset        for (let 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    let str = "abc";    let index = -1;    let curr = "";    powerSet(str, index, curr);              // This code is contributed by rag2127</script>
a
ab
abc
ac
b
bc
c
 
Method 2: 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.  
C++
Java
Python3
C#
Javascript
// 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, currvoid 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 codeint main(){    string str = "abc";    powerSet(str);    return 0;}
// Java program to generate power setclass GFG { // str : Stores input string// curr : Stores current subset// index : Index in current subset, currstatic 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 codepublic static void main(String[] args){    String str = "abc";        int index = 0;        String curr="";    powerSet(str,index,curr);     }}// This code is contributed by 29AjayKumar
# Python3 program to generate power setdef 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 Codeif __name__ == "__main__":     s1 = "abc"    index = 0    curr = ""    powerSet(s1, index , curr) # This code is contributed by Ekta Singh
// C# program to generate power setusing 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
<script>// Javascript program to generate power set         // str : Stores input string// curr : Stores current subset// index : Index in current subset, curr    function powerSet(str,index,curr)    {        let n = str.length;      // base case    if (index == n)    {        document.write(curr+"<br>");        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    let str = "abc";    let index = 0;    let curr="";    powerSet(str,index,curr);          // This code is contributed by avanitrachhadiya2155</script>
abc
ab
ac
a
bc
b
c
 
Method 3: 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. We’ll use ArrayList for this purpose For ex, f(0) = {a}, {} // {} when we don’t include any element from the set, it is null i.e {}. f(1) = {a}, {}, {b}, {a, b} // We have to copy all the elements from f(0) and then include the very next element from the set i.e b. So f(1) = f(0) + 1; f(2) = {a}, {}, {b}, {a, b}, {a, c}, {c}, {b, c}, {a, b, c}. So f(2) = f(1) +2;The general form becomes f(n) = f(n-1) + n; 
Java
// Java Recursive code to print// all subsets of set using ArrayListimport 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;    }}
[[], [a], [b], [a, b], , [a, c], [b, c], [a, b, c]]
 
Iterative program for power set. 
Anurag Vishwa
29AjayKumar
Rajput-Ji
princiraj1992
ukasp
rambabuy
SURENDRA_GANGWAR
ekta1991singh
rag2127
avanitrachhadiya2155
whoibrar
subset
Backtracking
Recursion
Recursion
subset
Backtracking
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