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Print all subsequences of a string

  • Difficulty Level : Medium
  • Last Updated : 25 Aug, 2021

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: 

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

Input : aaa
Output : a, aa, aaa

Method 1 (Pick and Don’t Pick Concept)  



C++




// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
 
// Find all subsequences
void printSubsequence(string input, string output)
{
    // Base Case
    // if the input is empty print the output string
    if (input.empty()) {
        cout << output << endl;
        return;
    }
 
    // output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.substr(1), output + input[0]);
 
    // output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.substr(1), output);
}
 
// Driver code
int main()
{
    // output is set to null before passing in as a
    // parameter
    string output = "";
    string input = "abcd";
 
    printSubsequence(input, output);
 
    return 0;
}
Java



// Java program for the above approach
import java.util.*;
class GFG {
 
    // Declare a global list
    static List<String> al = new ArrayList<>();
 
    // 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);
    }
}
Python3



# Below is the implementation of the above approach
def printSubsequence(input, output):
 
    # Base Case
    # if the input is empty print the output string
    if len(input) == 0:
        print(output, end=' ')
        return
 
    # output is passed with including the
    # 1st character of input string
    printSubsequence(input[1:], output+input[0])
 
    # output is passed without including the
    # 1st character of input string
    printSubsequence(input[1:], output)
 
 
# Driver code
# output is set to null before passing in
# as a parameter
output = ""
input = "abcd"
 
printSubsequence(input, output)
 
# This code is contributed by Tharun Reddy
C#



// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
static void printSubsequence(string input,
                             string output)
{
     
    // Base Case
    // If the input is empty print the output string
    if (input.Length == 0)
    {
        Console.WriteLine(output);
        return;
    }
 
    // Output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.Substring(1),
                     output + input[0]);
 
    // Output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.Substring(1),
                     output);
}
 
// Driver code
static void Main()
{
     
    // output is set to null before passing
    // in as a parameter
    string output = "";
    string input = "abcd";
     
    printSubsequence(input, output);
}
}
  
// This code is contributed by SoumikMondal
Javascript



<script>
 
// JavaScript program for the above approach
 
// Find all subsequences
function printSubsequence(input, output)
{
    // Base Case
    // if the input is empty print the output string
    if (input.length==0) {
        document.write( output + "<br>");
        return;
    }
 
    // output is passed with including
    // the Ist character of
    // Input string
    printSubsequence(input.substring(1), output + input[0]);
 
    // output is passed without
    // including the Ist character
    // of Input string
    printSubsequence(input.substring(1), output);
}
 
// Driver code
// output is set to null before passing in as a
// parameter
var output = "";
var input = "abcd";
printSubsequence(input, output);
 
 
</script>
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 substring 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 program 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;
 
// Function 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(); 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<>();
 
    // Function 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
aab aa aac bc b abc aabc ab ac a c 
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 the 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;
 
    if (!curr.empty()) {
        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;
        }
        if (curr != null && !curr.trim().isEmpty()) {
            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 write.geeksforgeeks.org or mail your article to review-team@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.
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