Print all subsequences of a string - GeeksforGeeks

06 Apr, 2022
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
Method 1 (Pick and Don’t Pick Concept)  
C++
Java
Python3
C#
Javascript
// C++ program for the above approach#include <bits/stdc++.h>using namespace std; // Find all subsequencesvoid 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 codeint main(){    // output is set to null before passing in as a    // parameter    string output = "";    string input = "abcd";     printSubsequence(input, output);     return 0;}
// Java program for the above approachimport 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);    }}
# Below is the implementation of the above approachdef 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 parameteroutput = ""input = "abcd" printSubsequence(input, output) # This code is contributed by Tharun Reddy
// C# program for the above approachusing 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 codestatic 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
<script> // JavaScript program for the above approach // Find all subsequencesfunction 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// parametervar output = "";var input = "abcd";printSubsequence(input, output);  </script>
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++
Java
Javascript
// CPP program to print all subsequence of a// given string.#include <bits/stdc++.h>using namespace std; // set to store all the subsequencesunordered_set<string> st; // Function computes all the subsequence of an stringvoid 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 Codeint main(){    string s = "aabc";    subsequence(s);    for (auto i : st)        cout << i << " ";    cout << endl;     return 0;} // This code is contributed by// sanjeev2552
// 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);    }}
<script> // JavaScript program to print all subsequence of a// given string.  // set to store all the subsequenceslet st = new Set(); // Function computes all the subsequence of an stringfunction 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>
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++
Java
Python3
Javascript
// 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, currvoid 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 codeint main(){    string str = "cab";    printSubSeq(str);    return 0;}
// 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
# 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, currdef 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.#  functiondef printSubSeq(str):   printSubSeqRec(str, len(str)) # // Driver codestr = "cab"printSubSeq(str) # This code is contributed by shinjanpatra
<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>
c
ca
cab
cb
a
ab
b
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