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


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.

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// 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);
    }
}

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Output:

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

Time complexity : O(n^3)

Alternate Solution :
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++

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// 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;
}

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Java

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// 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

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Output:

c
ca
cab
cb
a
ab
b

Time complexity : O(2^n)
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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