Print all subsequences of a string

Last Updated : 31 Jul, 2023

Given a string, we have to find out all its 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, a, a, aa, aa, aa, aaa

Method 1 (Pick and Don’t Pick Concept) :

Implementation:

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

Time Complexity: O(2^n)
The time complexity of this approach is O(2^n), where n is the length of the given string. This is because, for a string of length n, we are generating a total of 2^n subsequences.

Space Complexity: O(n)
The recursive function call stack requires O(n) space for the worst case, where n is the length of the given string.

Method 2:

Note: This method does not handle duplicate characters.

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);
    }
}

Python3

# Python program to print all subsequence of a
# given string.
 
# set to store all the subsequences
st = set()
 
# Function computes all the subsequence of an string
def subsequence(str):
 
    # Iterate over the entire string
    for i in range(len(str)):
 
        # Iterate from the end of the string
        # to generate substrings
        for j in range(len(str),i,-1):
            sub_str = str[i: i+j]
            st.add(sub_str)
 
            # Drop kth character in the substring
            # and if its not in the set then recur
            for k in range(1,len(sub_str)):
                sb = sub_str
 
                # Drop character from the string
                sb = sb.replace(sb[k],"")
                subsequence(sb)
 
# Driver Code
 
s = "aabc"
subsequence(s)
for i in st:
    print(i,end = " ")
print()
 
# This code is contributed by shinjanpatra

C#

using System;
using System.Collections.Generic;
using System.Text;
public class GFG {
 
  // set to store all the subsequences
  public static HashSet<string> st
    = new HashSet<string>();
   
  // Function computes all the subsequence of an string
  public 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 - i);
        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; k++) {
          string sb = sub_str;
          // Drop character from the string
          StringBuilder strBuilder
            = new StringBuilder(sb);
          strBuilder.Remove(k, 1);
          sb = strBuilder.ToString();
          subsequence(sb);
        }
      }
    }
  }
 
  static public void Main()
  {
 
    string s = "aabc";
    subsequence(s);
    foreach(var value in st)
    {
      Console.Write(value);
      Console.Write(" ");
    }
  }
}
 
// This code is contributed by akashish__

Javascript

<script>
 
// JavaScript program to print all subsequence of a
// given string.
 
 
// set to store all the subsequences
let st = new Set();
 
// Function computes all the subsequence of an string
function subsequence(str)
{
 
    // Iterate over the entire string
    for (let i = 0; i < str.length; i++) {
 
        // Iterate from the end of the string
        // to generate substrings
        for (let j = str.length; j > i; j--) {
            let sub_str = str.substr(i, i+j);
            st.add(sub_str);
 
            // Drop kth character in the substring
            // and if its not in the set then recur
            for (let k = 1; k < sub_str.length; k++) {
                let sb = sub_str;
 
                // Drop character from the string
                sb =sb.replace(sb[k],"");
                subsequence(sb);
            }
        }
    }
}
 
// Driver Code
 
let s = "aabc";
subsequence(s);
for (let i of st)
    document.write(i," ");
document.write("</br>");
 
// This code is contributed by shinjanpatra
 
</script>

Output

aab aa aac bc b abc aabc ab ac a c

Time Complexity: O(N^3)

Auxiliary Space: O(N)

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

Implementation:

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

Python3

# Python program to generate power set in lexicographic order.
 
 # str: Stores input string
 # n: Length of str.
 # curr: Stores current permutation
 # index: Index in current permutation, curr
def printSubSeqRec(str, n, index = -1, curr = ""):
   
  # base case
     if (index == n):
       return
     if (len(curr) > 0):
       print(curr)
 
     i = index + 1
 
     while(i < n):
        curr = curr + str[i]
        printSubSeqRec(str, n, i, curr)
        curr = curr[0:-1]
        i = i + 1
        
#  Generates power set in lexicographic order.
#  function
def printSubSeq(str):
   printSubSeqRec(str, len(str))
 
# // Driver code
str = "cab"
printSubSeq(str)
 
# This code is contributed by shinjanpatra

C#

// Include namespace system
using System;
 
// C# program to generate power set in
// lexicographic order.
public class GFG
{
    // str : Stores input string
    // n : Length of str.
    // curr : Stores current permutation
    // index : Index in current permutation, curr
    public static void printSubSeqRec(String str, int n, int index, String curr)
    {
        // base case
        if (index == n)
        {
            return;
        }
        if (curr != null && !(curr.Trim().Length == 0))
        {
            Console.WriteLine(curr);
        }
        for (int i = index + 1; i < n; i++)
        {
            curr += str[i];
            GFG.printSubSeqRec(str, n, i, curr);
            // backtracking
            curr = curr.Substring(0,curr.Length - 1-0);
        }
    }
    // Generates power set in
    // lexicographic order.
    public static void printSubSeq(String str)
    {
        var index = -1;
        var curr = "";
        GFG.printSubSeqRec(str, str.Length, index, curr);
    }
    // Driver code
    public static void Main(String[] args)
    {
        var str = "cab";
        GFG.printSubSeq(str);
    }
}
// This code is contributed by mukulsomukesh

Javascript

<script>
// JavaScript program to generate power set in
// lexicographic order.
  
// str : Stores input string
// n : Length of str.
// curr : Stores current permutation
// index : Index in current permutation, curr
 
function printSubSeqRec(str,n,index = -1,curr = "")
{
    // base case
    if (index == n)
        return;
  
    if (curr.length>0) {
        document.write(curr)
    }
  
    for (let i = index + 1; i < n; i++) {
  
        curr += str[i];
        printSubSeqRec(str, n, i, curr);
  
        // backtracking
        curr = curr.slice(0, - 1);
    }
    return;
}
  
// Generates power set in lexicographic
// order.
function printSubSeq(str)
{
    printSubSeqRec(str, str.length);
}
  
// Driver code
 
let str = "cab";
printSubSeq(str);
</script>

Output

c
ca
cab
cb
a
ab
b

Time Complexity: O(n * 2ⁿ), where n is the size of the given string
Auxiliary Space: O(n), due to recursive call stack

Method 4: Using Binary representation of numbers to create Subsequences

String = “abc”

All combinations of abc can be represented by all binary representation from 0 to (2^n – 1) where n is the size of the string . The following representation clears things up.

Note : We can also take zero into consideration which will eventually give us an empty set “” , the only change in code will be starting loop from zero.

001 -> “c”
010 -> “b”
011 -> “bc
100 -> “a”
101 -> “ac”
110 -> “ab”
111 -> “abc”

As you can observe we get unique sub-sequences for every set-bit and thus no 2 combinations can be same as 2 numbers cannot have same binary representation.

Input : “abc”
Output :
a
b
a b
c
a c
b c
a b c

Implementation :

We take the string as input.
We declare a vector of strings to store each sub-sequence as a string.
Each time call the function with 1,2,3 and so on and we only push those indexes in our string whose bit is set and we keep incrementing our index pointer.
Once we have generated the corresponding sub-sequence for a binary representation we can push this string into our vector of strings.
Finally, we can print our vector of strings and get the desired output.

Below is the implementation of the above approach:

C++

#include<bits/stdc++.h>
using namespace std;
 
string print(string s , int i){    
    int j = 0;
    string sub;
     
    //finding where the bit is set
    while(i>0){
        if(i & 1){
            sub.push_back(s[j]); //pushing only when bit is set 
        }
        j++;  //always incrementing the index pointer
        i = i >> 1;
    }
     
    return sub;
}
 
vector<string> createsubsets(string& s){
     
    vector <string> res;
    for(int i = 1 ; i <= ((1 << s.size()) - 1) ; i++){
        //each time we create a subsequence for corresponding binary representation
        res.push_back(print(s,i));
    }
    return res;
}
 
int main(){
    string s = "abc";    
    //vector of strings to store all sub-sequences
    vector <string> print = createsubsets(s);
     
    //print function
    for(int i = 0 ; i < print.size() ; i++){
        for (int j = 0; j < print[i].size(); j++){
            cout << print[i][j]<<" ";
        }
        cout << endl;
    }
    return 0;
}

Java

import java.util.*;
 
public class GFG {
     
    //function to find where the bit is set
    public static String print(String s , int i){
         
        int j = 0;
        String sub = "";
         
        //finding the bit is set
        while(i>0){
            if((i & 1) == 1){
                sub += s.charAt(j);
            }
            j++;
            i = i>>1;
        }
        return sub;
    }
     
    //function to create sub-sets
    public static List<String> createsubsets(String s){
         
        List<String> res = new ArrayList<>();
        for(int i = 0 ; i < (1<<s.length()) ; i++){
            //each time we create a subsequence for corresponding binary representation
            res.add(print(s,i));
        }
        return res;
    }
     
    //main function to call 
    public static void main(String args[]) 
    { 
        String s = "abc";
         
        // vector of strings to store all sub-sequences 
        List<String> print = createsubsets(s); 
   
        // print the subsets 
        for (int i = 0; i < print.size(); i++) { 
            for (int j = 0; j < print.get(i).length(); j++) { 
                System.out.print(print.get(i).charAt(j) + " "); 
            } 
            System.out.println(); 
        } 
    } 
}
// This code contributed by Srj_27

Python3

def print_subset(s, i):
    j = 0
    sub = ""
    #finding where the bit is set
    while i > 0:
        if i & 1:
            sub += s[j] #pushing only when bit is set 
        j += 1 #always incrementing the index pointer
        i = i >> 1
    return sub
 
def createsubsets(s):
    res = []
    for i in range(1, (1 << len(s))):
        #each time we create a subsequence for corresponding binary representation
        res.append(print_subset(s, i))
    return res
 
if __name__ == "__main__":
    s = "abc"
    #vector of strings to store all sub-sequences
    subsets = createsubsets(s)
     
    #print function
    for subset in subsets:
        for c in subset:
            print(c, end=" ")
        print()
 
# This code is contributed Shivam Tiwari

C#

using System;
using System.Collections.Generic;
 
namespace GFG
{
    class Program
    {
        //function to find where the bit is set
        public static string Print(string s, int i)
        {
            int j = 0;
            string sub = "";
 
            //finding the bit is set
            while (i > 0)
            {
                if ((i & 1) == 1)
                {
                    sub += s[j];
                }
                j++;
                i = i >> 1;
            }
            return sub;
        }
 
        //function to create sub-sets
        public static List<string> CreateSubsets(string s)
        {
            List<string> res = new List<string>();
            for (int i = 0; i < (1 << s.Length); i++)
            {
                //each time we create a subsequence for corresponding binary representation
                res.Add(Print(s, i));
            }
            return res;
        }
 
        static void Main(string[] args)
        {
            string s = "abc";
 
            // list of strings to store all sub-sequences 
            List<string> print = CreateSubsets(s);
 
            // print the subsets 
            for (int i = 0; i < print.Count; i++)
            {
                for (int j = 0; j < print[i].Length; j++)
                {
                    Console.Write(print[i][j] + " ");
                }
                Console.WriteLine();
            }
        }
    }
}
// This code contributed by Ajax

Javascript

// Function to extract a subsequence when the corresponding bit is set
function printSubset(s, i) {
  let j = 0; // Index pointer to iterate through the string s
  let sub = ""; // Resultant subsequence
   
  // Finding where the bit is set
  while (i > 0) {
    if (i & 1) {
      // Pushing only when the bit is set
      sub += s[j]; 
    }
    j += 1; // Always incrementing the index pointer
    i = i >> 1; // Right shift the number to get the next bit
  }
  return sub;
}
 
// Function to generate all possible sub-sequences
function createSubsets(s) {
  let res = []; // Array to store all sub-sequences
  for (let i = 1; i < (1 << s.length); i++) {
    // Each iteration generates a sub-sequence for the corresponding binary representation
    res.push(printSubset(s, i));
  }
  return res;
}
 
 
// Driver Code
const s = "abc"; // String input
const subsets = createSubsets(s); // Array of all sub-sequences
 
// Printing the sub-sequences
for (let subset of subsets) {
  for (let c of subset) {
    process.stdout.write(c + " ");
  }
  console.log();
}

Output

a 
b 
a b 
c 
a c 
b c 
a b c

Strings in different language

Basic operations on String

Binary String

Substring and Subsequence

Palindrome

Easy problems on String

Intermediate problems on String

Hard problems on String

Strings in different language

Basic operations on String

Binary String

Substring and Subsequence

Palindrome

Easy problems on String

Intermediate problems on String

Hard problems on String

Print all subsequences of a string

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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