Print all subsequences of a string

Given a string, we have to find out all subsequences of it. A String is a subsequence of a given String, that is generated by deleting some character of a given string without changing its order.

Examples:

Input : abc
Output : a, b, c, ab, bc, ac, abc

Input : aaa
Output : a, aa, aaa

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Pick and Don’t Pick Concept)

import java.util.*; 
class GFG{ 
  
        // creating a public static Arraylist such that 
        // we can store values 
    // IF there is any question of returning the  
        // we can directly return too// 
    public static ArrayList<String>al=new ArrayList<String>(); 
    public static void main(String[] args) { 
        String s="abcd"; 
        findsubsequences(s,"");    // Calling a function 
        System.out.println(al); 
    } 
  
    private static void findsubsequences(String s, String ans) { 
        if(s.length()==0) 
        { 
            al.add(ans);  
            return; 
        } 
  
                //we add adding 1st character in string            
        findsubsequences(s.substring(1),ans+s.charAt(0)) ; 
  
                // Not adding first character of the string 
                // because the concept of subsequence either  
                // character will present or not 
        findsubsequences(s.substring(1),ans);       
    } 
} 

Output:

[abcd, abc, abd, ab, acd, ac, ad, a, bcd, bc, bd, b, cd, c, d, ]

Method 2
Explanation :

Step 1: Iterate over the entire String
Step 2: Iterate from the end of string 
        in order to generate different substring
        add the subtring to the list
Step 3: Drop kth character from the substring obtained 
        from above to generate different subsequence.
Step 4: if the subsequence is not in the list then recur.

Below is the implementation of the approach.

C++

// CPP rogram to print all subsequence of a 
// given string. 
#include <bits/stdc++.h> 
using namespace std; 
  
// set to store all the subsequences 
unordered_set<string> st; 
  
// It computes all the subsequence of an string 
void subsequence(string str) 
{ 
    // iterate over the entire string 
    for (int i = 0; i < str.length(); i++) 
    { 
        // iterate from the end of the string 
        // to generate substrings 
        for (int j = str.length(); j > i; j--) 
        { 
            string sub_str = str.substr(i, j); 
            st.insert(sub_str); 
  
            // drop kth character in the substring 
            // and if its not in the set then recur 
            for (int k = 1; k < sub_str.length() - 1; k++) 
            { 
                string sb = sub_str; 
  
                // drop character from the string 
                sb.erase(sb.begin() + k); 
                subsequence(sb); 
            } 
        } 
    } 
} 
  
// Driver Code 
int main() 
{ 
    string s = "aabc"; 
    subsequence(s); 
    for (auto i : st) 
        cout << i << " "; 
    cout << endl; 
  
    return 0; 
} 
  
// This code is contributed by 
// sanjeev2552 

Java

// Java Program to print all subsequence of a 
// given string. 
import java.util.HashSet; 
  
public class Subsequence { 
      
    // set to store all the subsequences 
    static HashSet<String> st = new HashSet<>(); 
  
    // It computes all the subsequence of an string 
    static void subsequence(String str) 
    { 
        // iterate over the entire string 
        for (int i = 0; i < str.length(); i++) { 
              
            // iterate from the end of the string 
            // to generate substrings 
            for (int j = str.length(); j > i; j--) { 
                String sub_str = str.substring(i, j); 
              
                if (!st.contains(sub_str)) 
                    st.add(sub_str); 
  
                // drop kth character in the substring 
                // and if its not in the set then recur 
                for (int k = 1; k < sub_str.length() - 1; k++) { 
                    StringBuffer sb = new StringBuffer(sub_str); 
  
                    // drop character from the string 
                    sb.deleteCharAt(k); 
                    if (!st.contains(sb)) 
                        ; 
                    subsequence(sb.toString()); 
                } 
            } 
        } 
    } 
  
    // Driver code 
    public static void main(String[] args) 
    { 
        String s = "aabc"; 
        subsequence(s); 
        System.out.println(st); 
    } 
} 

Output:

[aa, a, ab, bc, ac, b, aac, abc, c, aab, aabc]

Method 3 :
One by one fix characters and recursively generates all subsets starting from them. After every recursive call, we remove last character so that next permutation can be generated.

C++

// CPP program to generate power set in 
// lexicographic order. 
#include <bits/stdc++.h> 
using namespace std; 
   
// str : Stores input string 
// n : Length of str. 
// curr : Stores current permutation 
// index : Index in current permutation, curr 
void printSubSeqRec(string str, int n, 
           int index = -1, string curr = "") 
{ 
    // base case 
    if (index == n)  
        return; 
   
    cout << curr << "\n"; 
    for (int i = index + 1; i < n; i++) { 
   
        curr += str[i]; 
        printSubSeqRec(str, n, i, curr); 
    
        // backtracking 
        curr = curr.erase(curr.size() - 1);  
    } 
    return; 
} 
   
// Generates power set in lexicographic 
// order. 
void printSubSeq(string str) 
{ 
    printSubSeqRec(str, str.size()); 
} 
   
// Driver code 
int main() 
{ 
    string str = "cab"; 
    printSubSeq(str); 
    return 0; 
} 

Java

// Java program to generate power set in  
// lexicographic order. 
class GFG  
{ 
  
    // str : Stores input string  
    // n : Length of str.  
    // curr : Stores current permutation  
    // index : Index in current permutation, curr  
    static void printSubSeqRec(String str, int n, 
                            int index, String curr)  
    { 
        // base case  
        if (index == n)  
        { 
            return; 
        } 
        System.out.println(curr); 
        for (int i = index + 1; i < n; i++)  
        { 
            curr += str.charAt(i); 
            printSubSeqRec(str, n, i, curr); 
  
            // backtracking  
            curr = curr.substring(0, curr.length() - 1); 
        } 
    } 
  
    // Generates power set in   
    // lexicographic order.  
    static void printSubSeq(String str)  
    { 
        int index = -1; 
        String curr = ""; 
  
        printSubSeqRec(str, str.length(), index, curr); 
    } 
  
    // Driver code  
    public static void main(String[] args)  
    { 
        String str = "cab"; 
        printSubSeq(str); 
    } 
}  
  
// This code is contributed by PrinciRaj1992 

Output:

c
ca
cab
cb
a
ab
b

This article is contributed by Sumit Ghosh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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My Personal Notes

Recommended: Please try your approach on {IDE} first, before moving on to the solution.

C++

Java

C++

Java

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