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++
// CPP 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 characther 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); } } |
[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 subtring 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 rogram 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() - 1; 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); } } |
Output:
[aa, a, ab, bc, ac, b, aac, abc, c, aab, aabc]
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++
// 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 |
Output:
c ca cab cb a ab b
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