# Print all subsequences of a string

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
```

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

Method 1 (Pick and Don’t Pick Concept)

 `import` `java.util.*; ` `class` `GFG{ ` ` `  `        ``// 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` `ArrayListal=``new` `ArrayList(); ` `    ``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);       ` `    ``} ` `} `

Output:

```[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 ` `using` `namespace` `std; ` ` `  `// set to store all the subsequences ` `unordered_set st; ` ` `  `// It 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 st = ``new` `HashSet<>(); ` ` `  `    ``// It 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 next permutation can be generated.

## C++

 `// CPP program to generate power set in ` `// lexicographic order. ` `#include ` `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``; ` `  `  `    ``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``; ` `        ``} ` `        ``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
```

This article is contributed by Sumit Ghosh. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

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