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

**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.

`// 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<>(); ` ` ` ` ` `// 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); ` ` ` `} ` `} ` |

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Output:

[aa, a, ab, bc, ac, b, aac, abc, c, aab, aabc]

**Alternate Solution : **

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.

`// 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` `; ` ` ` ` ` `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; ` `} ` |

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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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