Print all subsequences of a string | Iterative Method

Last Updated : 20 Feb, 2023

Given a string s, print all possible subsequences of the given string in an iterative manner. We have already discussed Recursive method to print all subsequences of a string.

Examples:

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

Approach 1 :
Here, we discuss much easier and simpler iterative approach which is similar to Power Set. We use bit pattern from binary representation of 1 to 2^length(s) – 1.

input = “abc”
Binary representation to consider 1 to (2^3-1), i.e 1 to 7.
Start from left (MSB) to right (LSB) of binary representation and append characters from input string which corresponds to bit value 1 in binary representation to Final subsequence string sub.

Example:

001 => abc . Only c corresponds to bit 1. So, subsequence = c. 101 => abc . a and c corresponds to bit 1. So, subsequence = ac. binary_representation (1) = 001 => c binary_representation (2) = 010 => b binary_representation (3) = 011 => bc binary_representation (4) = 100 => a binary_representation (5) = 101 => ac binary_representation (6) = 110 => ab binary_representation (7) = 111 => abc

Below is the implementation of above approach:

C++

// C++ program to print all Subsequences
// of a string in iterative manner
#include <bits/stdc++.h>
using namespace std;
 
// function to find subsequence
string subsequence(string s, int binary, int len)
{
    string sub = "";
    for (int j = 0; j < len; j++)
 
        // check if jth bit in binary is 1
        if (binary & (1 << j))
 
            // if jth bit is 1, include it
            // in subsequence
            sub += s[j];
 
    return sub;
}
 
// function to print all subsequences
void possibleSubsequences(string s){
 
    // map to store subsequence 
    // lexicographically by length
    map<int, set<string> > sorted_subsequence;
 
    int len = s.size();
     
    // Total number of non-empty subsequence
    // in string is 2^len-1
    int limit = pow(2, len);
     
    // i=0, corresponds to empty subsequence
    for (int i = 1; i <= limit - 1; i++) { 
         
        // subsequence for binary pattern i
        string sub = subsequence(s, i, len);
         
        // storing sub in map
        sorted_subsequence[sub.length()].insert(sub);
    }
 
    for (auto it : sorted_subsequence) { 
         
        // it.first is length of Subsequence
        // it.second is set<string>
        cout << "Subsequences of length = "
             << it.first << " are:" << endl;
              
        for (auto ii : it.second)
             
            // ii is iterator of type set<string>
            cout << ii << " ";
         
        cout << endl;
    }
}
 
// driver function
int main()
{
    string s = "aabc";
    possibleSubsequences(s);
    return 0;
}

Java

// Java program to print all Subsequences
// of a String in iterative manner
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Map;
import java.util.SortedMap;
import java.util.TreeMap;
 
class Graph{
 
// Function to find subsequence
static String subsequence(String s, 
                          int binary,
                          int len) 
{
    String sub = "";
     
    for(int j = 0; j < len; j++)
     
        // Check if jth bit in binary is 1
        if ((binary & (1 << j)) != 0)
 
            // If jth bit is 1, include it
            // in subsequence
            sub += s.charAt(j);
 
    return sub;
}
 
// Function to print all subsequences
static void possibleSubsequences(String s)
{
     
    // Map to store subsequence
    // lexicographically by length
    SortedMap<Integer, 
              HashSet<String>> sorted_subsequence = new TreeMap<Integer,
                                                                HashSet<String>>();
 
    int len = s.length();
 
    // Total number of non-empty subsequence
    // in String is 2^len-1
    int limit = (int) Math.pow(2, len);
 
    // i=0, corresponds to empty subsequence
    for(int i = 1; i <= limit - 1; i++) 
    {
         
        // Subsequence for binary pattern i
        String sub = subsequence(s, i, len);
         
        // Storing sub in map
        if (!sorted_subsequence.containsKey(sub.length()))
            sorted_subsequence.put(
                sub.length(), new HashSet<>());
            sorted_subsequence.get(
                sub.length()).add(sub);
    }
 
    for(Map.Entry<Integer,
                  HashSet<String>> it : sorted_subsequence.entrySet()) 
    {
         
        // it.first is length of Subsequence
        // it.second is set<String>
        System.out.println("Subsequences of length = " + 
                           it.getKey() + " are:");
                            
        for(String ii : it.getValue())
         
            // ii is iterator of type set<String>
            System.out.print(ii + " ");
 
        System.out.println();
    }
}
 
// Driver code
public static void main(String[] args)
{
    String s = "aabc";
     
    possibleSubsequences(s);
}
}
 
// This code is contributed by sanjeev2552

Python3

# Python3 program to print all Subsequences
# of a string in iterative manner
 
# function to find subsequence
def subsequence(s, binary, length):
    sub = ""
    for j in range(length):
         
        # check if jth bit in binary is 1
        if (binary & (1 << j)):
 
            # if jth bit is 1, include it
            # in subsequence
            sub += s[j]
    return sub
 
# function to print all subsequences
def possibleSubsequences(s):
 
    # map to store subsequence 
    # lexicographically by length
    sorted_subsequence = {}
 
    length = len(s)
     
    # Total number of non-empty subsequence
    # in string is 2^len-1
    limit = 2 ** length
     
    # i=0, corresponds to empty subsequence
    for i in range(1, limit):
         
        # subsequence for binary pattern i
        sub = subsequence(s, i, length)
         
        # storing sub in map
        if len(sub) in sorted_subsequence.keys():
            sorted_subsequence[len(sub)] = \
             tuple(list(sorted_subsequence[len(sub)]) + [sub])
        else:
            sorted_subsequence[len(sub)] = [sub]
 
    for it in sorted_subsequence:
         
        # it.first is length of Subsequence
        # it.second is set<string>
        print("Subsequences of length =", it, "are:")
        for ii in sorted(set(sorted_subsequence[it])):
             
            # ii is iterator of type set<string>
            print(ii, end = ' ')
        print()
 
# Driver Code
s = "aabc"
possibleSubsequences(s)
 
# This code is contributed by ankush_953

C#

// C# program to print all Subsequences
// of a String in iterative manner
using System;
using System.Collections.Generic;
 
class Graph {
 
  // Function to find subsequence
  static string subsequence(string s, int binary, int len)
  {
    string sub = "";
 
    for (int j = 0; j < len; j++)
 
      // Check if jth bit in binary is 1
      if ((binary & (1 << j)) != 0)
 
        // If jth bit is 1, include it
        // in subsequence
        sub += s[j];
 
    return sub;
  }
 
  // Function to print all subsequences
  static void possibleSubsequences(string s)
  {
 
    // Map to store subsequence
    // lexicographically by length
    SortedDictionary<int, HashSet<string> >
      sorted_subsequence
      = new SortedDictionary<int, HashSet<string> >();
 
    int len = s.Length;
 
    // Total number of non-empty subsequence
    // in String is 2^len-1
    int limit = (int)Math.Pow(2, len);
 
    // i=0, corresponds to empty subsequence
    for (int i = 1; i <= limit - 1; i++) {
 
      // Subsequence for binary pattern i
      string sub = subsequence(s, i, len);
 
      // Storing sub in map
      if (!sorted_subsequence.ContainsKey(sub.Length))
        sorted_subsequence[sub.Length]
        = new HashSet<string>();
      sorted_subsequence[sub.Length].Add(sub);
    }
 
    foreach(var it in sorted_subsequence)
    {
 
      // it.first is length of Subsequence
      // it.second is set<String>
      Console.WriteLine("Subsequences of length = "
                        + it.Key + " are:");
 
      foreach(String ii in it.Value)
 
        // ii is iterator of type set<String>
        Console.Write(ii + " ");
 
      Console.WriteLine();
    }
  }
 
  // Driver code
  public static void Main(string[] args)
  {
    string s = "aabc";
 
    possibleSubsequences(s);
  }
}
 
// This code is contributed by phasing17

Javascript

<script>
 
// Javascript program to print all Subsequences
// of a string in iterative manner
 
// function to find subsequence
function subsequence(s, binary, len)
{
    let sub = "";
    for(let j = 0; j < len; j++)
     
        // Check if jth bit in binary is 1
        if (binary & (1 << j))
 
            // If jth bit is 1, include it
            // in subsequence
            sub += s[j];
 
    return sub;
}
 
// Function to print all subsequences
function possibleSubsequences(s)
{
     
    // map to store subsequence
    // lexicographically by length
    let sorted_subsequence = new Map();
 
    let len = s.length;
 
    // Total number of non-empty subsequence
    // in string is 2^len-1
    let limit = Math.pow(2, len);
 
    // i=0, corresponds to empty subsequence
    for(let i = 1; i <= limit - 1; i++)
    {
         
        // Subsequence for binary pattern i
        let sub = subsequence(s, i, len);
 
        // Storing sub in map
        if (!sorted_subsequence.has(sub.length))
            sorted_subsequence.set(sub.length, new Set());
             
        sorted_subsequence.get(sub.length).add(sub);
    }
 
    for(let it of sorted_subsequence) 
    {
         
        // it.first is length of Subsequence
        // it.second is set<string>
        document.write("Subsequences of length = " +
                       it[0] + " are:" + "<br>");
 
        for(let ii of it[1])
 
            // ii is iterator of type set<string>
            document.write(ii + " ");
 
        document.write("<br>");
    }
}
 
// Driver code
let s = "aabc";
 
possibleSubsequences(s);
 
// This code is contributed by gfgking
 
</script>

Output

Subsequences of length = 1 are: a b c  Subsequences of length = 2 are: aa ab ac bc  Subsequences of length = 3 are: aab aac abc  Subsequences of length = 4 are: aabc

Time Complexity : O(2^n), where n is length of string to find subsequences and n is length of binary string.
Space Complexity: O(1)

Approach 2 :

Approach is to get the position of rightmost set bit and reset that bit after appending corresponding character from given string to the subsequence and will repeat the same thing till corresponding binary pattern has no set bits.

If input is s = “abc” Binary representation to consider 1 to (2^3-1), i.e 1 to 7. 001 => abc . Only c corresponds to bit 1. So, subsequence = c 101 => abc . a and c corresponds to bit 1. So, subsequence = ac. Let us use Binary representation of 5, i.e 101. Rightmost bit is at position 1, append character at beginning of sub = c ,reset position 1 => 100 Rightmost bit is at position 3, append character at beginning of sub = ac ,reset position 3 => 000 As now we have no set bit left, we stop computing subsequence.

Example:

binary_representation (1) = 001 => c binary_representation (2) = 010 => b binary_representation (3) = 011 => bc binary_representation (4) = 100 => a binary_representation (5) = 101 => ac binary_representation (6) = 110 => ab binary_representation (7) = 111 => abc

Below is the implementation of above approach :

C++

// C++ code all Subsequences of a
// string in iterative manner
#include <bits/stdc++.h>
using namespace std;
 
// function to find subsequence
string subsequence(string s, int binary)
{
    string sub = "";
    int pos;
     
    // loop while binary is greater than 0
    while(binary>0)
    { 
        // get the position of rightmost set bit
        pos=log2(binary&-binary)+1;
         
        // append at beginning as we are 
        // going from LSB to MSB
        sub=s[pos-1]+sub;
         
        // resets bit at pos in binary
        binary= (binary & ~(1 << (pos-1)));
    }
    reverse(sub.begin(),sub.end());
    return sub;
}
 
// function to print all subsequences
void possibleSubsequences(string s){
 
    // map to store subsequence 
    // lexicographically by length
    map<int, set<string> > sorted_subsequence;
 
    int len = s.size();
     
    // Total number of non-empty subsequence
    // in string is 2^len-1
    int limit = pow(2, len);
     
    // i=0, corresponds to empty subsequence
    for (int i = 1; i <= limit - 1; i++) { 
         
        // subsequence for binary pattern i
        string sub = subsequence(s, i);
         
        // storing sub in map
        sorted_subsequence[sub.length()].insert(sub);
    }
 
    for (auto it : sorted_subsequence) { 
         
        // it.first is length of Subsequence
        // it.second is set<string>
        cout << "Subsequences of length = "
            << it.first << " are:" << endl;
             
        for (auto ii : it.second)
             
            // ii is iterator of type set<string>
            cout << ii << " ";
         
        cout << endl;
    }
}
 
// driver function
int main()
{
    string s = "aabc";
    possibleSubsequences(s);
     
    return 0;
}

Java

import java.util.*;
 
public class GFG {
    // function to find subsequence
    public static String subsequence(String s, int binary) {
        String sub = "";
        int pos;
 
        // loop while binary is greater than 0
        while (binary > 0) {
            // get the position of rightmost set bit
            pos = (int) (Math.log(binary & -binary) / Math.log(2)) + 1;
 
            // append at beginning as we are
            // going from LSB to MSB
            sub = s.charAt(pos - 1) + sub;
 
            // resets bit at pos in binary
            binary = (binary & ~(1 << (pos - 1)));
        }
        sub = new StringBuilder(sub).reverse().toString();
        return sub;
    }
 
    // function to print all subsequences
    public static void possibleSubsequences(String s) {
        // map to store subsequence
        // lexicographically by length
        Map<Integer, Set<String>> sortedSubsequence = new HashMap<>();
 
        int len = s.length();
 
        // Total number of non-empty subsequence
        // in string is 2^len-1
        int limit = (int) Math.pow(2, len);
 
        // i=0, corresponds to empty subsequence
        for (int i = 1; i <= limit - 1; i++) {
            // subsequence for binary pattern i
            String sub = subsequence(s, i);
 
            // storing sub in map
            sortedSubsequence.computeIfAbsent(sub.length(), k -> new TreeSet<>()).add(sub);
        }
 
        for (Map.Entry<Integer, Set<String>> it : sortedSubsequence.entrySet()) {
            // it.getKey() is length of Subsequence
            // it.getValue() is Set<String>
            System.out.println("Subsequences of length = " + it.getKey() + " are:");
 
            for (String ii : it.getValue())
                // ii is element of Set<String>
                System.out.print(ii + " ");
 
            System.out.println();
        }
    }
 
    // driver function
    public static void main(String[] args) {
        String s = "aabc";
        possibleSubsequences(s);
    }
}

Python3

# Python3 program to print all Subsequences
# of a string in an iterative manner
from math import log2, floor
 
# function to find subsequence
def subsequence(s, binary):
    sub = ""
     
    # loop while binary is greater than 
    while(binary > 0):
         
        # get the position of rightmost set bit
        pos=floor(log2(binary&-binary) + 1)
         
        # append at beginning as we are 
        # going from LSB to MSB
        sub = s[pos - 1] + sub
         
        # resets bit at pos in binary
        binary= (binary & ~(1 << (pos - 1)))
 
    sub = sub[::-1]
    return sub
 
# function to print all subsequences
def possibleSubsequences(s):
 
    # map to store subsequence 
    # lexicographically by length
    sorted_subsequence = {}
 
    length = len(s)
     
    # Total number of non-empty subsequence
    # in string is 2^len-1
    limit = 2 ** length
     
    # i=0, corresponds to empty subsequence
    for i in range(1, limit):
         
        # subsequence for binary pattern i
        sub = subsequence(s, i)
         
        # storing sub in map
        if len(sub) in sorted_subsequence.keys():
            sorted_subsequence[len(sub)] = \
            tuple(list(sorted_subsequence[len(sub)]) + [sub])
        else:
            sorted_subsequence[len(sub)] = [sub]
 
    for it in sorted_subsequence:
         
        # it.first is length of Subsequence
        # it.second is set<string>
        print("Subsequences of length =", it, "are:")
        for ii in sorted(set(sorted_subsequence[it])):
             
            # ii is iterator of type set<string>
            print(ii, end = ' ')
        print()
 
# Driver Code
s = "aabc"
possibleSubsequences(s)
 
# This code is contributed by ankush_953

C#

// C# program to print all Subsequences
// of a string in an iterative manner
 
using System;
using System.Collections.Generic;
using System.Linq;
 
class GFG 
{
 
  // function to find subsequence
  static string subsequence(string s, int binary)
  {
    string sub = "";
 
    // loop while binary is greater than
    while (binary > 0) {
 
      // get the position of rightmost set bit
      var pos = (int)Math.Floor(
        Math.Log(binary & (-binary))
        / Math.Log(2))
        + 1;
 
      // append at beginning as we are
      // going from LSB to MSB
      sub = s[pos - 1] + sub;
 
      // resets bit at pos in binary
      binary = (binary & ~(1 << (pos - 1)));
    }
 
    char[] charArray = sub.ToCharArray();
    Array.Reverse(charArray);
    return new string(charArray);
  }
 
  // function to print all subsequences
  static void possibleSubsequences(string s)
  {
 
    // map to store subsequence
    // lexicographically by length
    Dictionary<int, HashSet<string> > sorted_subsequence
      = new Dictionary<int, HashSet<string> >();
 
    int length = s.Length;
 
    // Total number of non-empty subsequence
    // in string is 2^len-1
    int limit = (int)Math.Pow(2, length);
 
    // i=0, corresponds to empty subsequence
    for (int i = 1; i < limit; i++) {
      // subsequence for binary pattern i
      string sub = subsequence(s, i);
 
      // storing sub in map
      if (sorted_subsequence.ContainsKey(
        sub.Length)) {
        sorted_subsequence[sub.Length].Add(sub);
      }
      else {
        sorted_subsequence[sub.Length]
          = new HashSet<string>();
        sorted_subsequence[sub.Length].Add(sub);
      }
    }
    foreach(var it in sorted_subsequence)
    {
 
      // it.first is length of Subsequence
      // it.second is set<string>
      Console.WriteLine("Subsequences of length = "
                        + it.Key + " are:");
      var arr = (it.Value).ToArray();
      Array.Sort(arr);
      for (int i = 0; i < arr.Length; i++) {
        Console.Write(arr[i] + " ");
      }
      Console.WriteLine();
    }
  }
 
  // Driver Code
  public static void Main(string[] args)
  {
    string s = "aabc";
    possibleSubsequences(s);
  }
}
 
// This code is contributed by phasing17

Javascript

// JavaScript program to print all Subsequences
// of a string in an iterative manner
 
// function to find subsequence
function subsequence(s, binary)
{
    var sub = "";
     
    // loop while binary is greater than 
    while(binary > 0)
    {
         
        // get the position of rightmost set bit
        var pos= Math.floor(Math.log2(binary&-binary) + 1);
         
        // append at beginning as we are 
        // going from LSB to MSB
        sub = s[pos - 1] + sub;
         
        // resets bit at pos in binary
        binary= (binary & ~(1 << (pos - 1)));
    }
 
    sub =  sub.split("");
    sub = sub.reverse();
    return sub.join("");
}
 
// function to print all subsequences
function possibleSubsequences(s)
{
 
    // map to store subsequence 
    // lexicographically by length
    var sorted_subsequence = {};
 
    var length = s.length;
     
    // Total number of non-empty subsequence
    // in string is 2^len-1
    var limit = 2 ** length;
     
    // i=0, corresponds to empty subsequence
    for (var i = 1; i < limit; i++)
    {
        // subsequence for binary pattern i
        var sub = subsequence(s, i);
         
        // storing sub in map
        if (sorted_subsequence.hasOwnProperty(sub.length))
        {
            //var list = sorted_subsequence[sub.length];
            //list.push(sub);
            sorted_subsequence[sub.length].push(sub);
        }
        else
            sorted_subsequence[sub.length] = [sub];
    }
    for (const it in sorted_subsequence)
    {
         
        // it.first is length of Subsequence
        // it.second is set<string>
        console.log("Subsequences of length =", it, "are:");
        var arr = sorted_subsequence[it];
        arr.sort();
        var set = new Set(arr);
        for (const ii of set)
            // ii is iterator of type set<string>
            process.stdout.write(ii + " ");
        console.log()
    }
}
 
// Driver Code
var s = "aabc";
possibleSubsequences(s);
 
// This code is contributed by phasing17

Output

Subsequences of length = 1 are: a b c  Subsequences of length = 2 are: aa ab ac bc  Subsequences of length = 3 are: aab aac abc  Subsequences of length = 4 are: aabc

Time Complexity: [Tex]O(2^{n} * b) [/Tex], where n is the length of string to find subsequence and b is the number of set bits in binary string.
Auxiliary Space: O(n)

Print all subsequences of a string using ArrayList

Abhishek rajput

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Print all subsequences of a string | Iterative Method

C++

Java

Python3

C#

Javascript

C++

Java

Python3

C#

Javascript

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