Print all subsequences of a string
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, abcInput : 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> |
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> |
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> |
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(); } |
a b a b c a c b c a b c
Time Complexity: O(n* 2^n)
Auxiliary Space: O(n)
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