# Generating distinct subsequences of a given string in lexicographic order

Given a string s, make a list of all possible combinations of letters of a given string S. If there are two strings with the same set of characters, print the lexicographically smallest arrangement of the two strings
For string abc, the list in lexicographic order subsequences are, a ab abc ac b bc c

Examples:

Input : s = "ab"
Output : a ab b

Input  : xyzx
Output : x xx xy xyx xyz xyzx xz xzx y
yx yz yzx z zx

The idea is to use a set (which is implemented using self balancing BST) to store subsequences so that duplicates can be tested.
To generate all subsequences, we one by one remove every character and recur for remaining string.

Implementation:

## C++

 // C++ program to print all distinct subsequences // of a string. #include using namespace std;   // Finds and stores result in st for a given // string s. void generate(set& st, string s) {     if (s.size() == 0)         return;       // If current string is not already present.     if (st.find(s) == st.end()) {         st.insert(s);           // Traverse current string, one by one         // remove every character and recur.         for (int i = 0; i < s.size(); i++) {             string t = s;             t.erase(i, 1);             generate(st, t);         }     }     return; }   // Driver code int main() {     string s = "xyz";     set st;     set::iterator it;     generate(st, s);     for (auto it = st.begin(); it != st.end(); it++)         cout << *it << endl;     return 0; }

## Java

 // Java program to print all distinct subsequences // of a string. import java.util.*;   class GFG {       // Finds and stores result in st for a given     // string s.     static void generate(Set st, String s)     {         if (s.length() == 0) {             return;         }           // If current string is not already present.         if (!st.contains(s)) {             st.add(s);               // Traverse current string, one by one             // remove every character and recur.             for (int i = 0; i < s.length(); i++) {                 String t = s;                 t = t.substring(0, i) + t.substring(i + 1);                 generate(st, t);             }         }         return;     }       // Driver code     public static void main(String args[])     {         String s = "xyz";         TreeSet st = new TreeSet<>();         generate(st, s);         for (String str : st) {             System.out.println(str);         }     } }   // This code has been contributed by 29AjayKumar // modified by rahul_107

## Python 3

 # Python program to print all distinct # subsequences of a string.   # Finds and stores result in st for a given # string s. def generate(st, s):     if len(s) == 0:         return       # If current string is not already present.     if s not in st:         st.add(s)           # Traverse current string, one by one         # remove every character and recur.         for i in range(len(s)):             t = list(s).copy()             t.remove(s[i])             t = ''.join(t)             generate(st, t)       return     # Driver Code if __name__ == "__main__":     s = "xyz"     st = set()     generate(st, s)     for i in st:         print(i)   # This code is contributed by # sanjeev2552

## C#

 // C# program to print all distinct subsequences // of a string. using System; using System.Collections.Generic;   class GFG {       // Finds and stores result in st for a given     // string s.     static void generate(HashSet st, String s)     {         if (s.Length == 0) {             return;         }           // If current string is not already present.         if (!st.Contains(s)) {             st.Add(s);               // Traverse current string, one by one             // remove every character and recur.             for (int i = 0; i < s.Length; i++) {                 String t = s;                 t = t.Substring(0, i) + t.Substring(i + 1);                 generate(st, t);             }         }         return;     }       // Driver code     public static void Main(String[] args)     {         String s = "xyz";         HashSet st = new HashSet();         generate(st, s);         foreach(String str in st)         {             Console.WriteLine(str);         }     } }   /* This code contributed by PrinciRaj1992 */

## Javascript



Output

x
xy
xyz
xz
y
yz
z

Time Complexity: O(nn)
Auxiliary Space: O(1)

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