# Count of non-empty sequences of a String

Given a string s, the task is to find the number of possible non-empty sequences of letters that can be made.

Examples:

```Input: "AAB"
Output: 8
Explanation:
1) A
2) AA
3) AAB
4) AB
5) ABA
6) B
7) BA
8) BAA
Total 8 possibilities

Input: "AAABBC"
Output: 188
```

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

Approach:
There are two possibilities either take the current character to our answer or leave it. We can solve this problem
To check for duplicates we can take set as a data Structure and will put our answers there and our count will be the size of our set.

Below is the implementation of the above approach:

## CPP

 `// C++ program for ` `// the above approach ` `#include ` `using` `namespace` `std; ` ` `  `// Recursive function which will ` `// calculate all the possibilities ` `// recursively ` `void` `recurr(string& tiles, vector<``bool``> vis, string ans, ` `            ``set& se) ` `{ ` `    ``if` `(ans.size() > 0) { ` `        ``// Check that the string ` `        ``// is already there or not ` `        ``if` `(se.count(ans)) ` `            ``return``; ` `        ``// Else put in set ` `        ``se.insert(ans); ` `    ``} ` `    ``// Run for all the ` `    ``// possibilities ` `    ``for` `(``int` `i = 0; i < tiles.size(); i++) { ` `        ``// If already taken ` `        ``// then don't do anything ` `        ``if` `(vis[i]) ` `            ``continue``; ` `        ``vis[i] = ``true``; ` `        ``// Else take it and ` `        ``// call recurr function ` `        ``recurr(tiles, vis, ans + tiles[i], se); ` `        ``vis[i] = ``false``; ` `    ``} ` `} ` ` `  `// Driver code ` `int` `main() ` `{ ` ` `  `    ``string s = ``"AAABBC"``; ` `    ``string curr = ``""``; ` ` `  `    ``set se; ` `    ``vector<``bool``> vis(s.size(), ``false``); ` `    ``recurr(s, vis, curr, se); ` ` `  `    ``int` `ans = se.size(); ` `    ``cout << ans << ``'\n'``; ` ` `  `    ``return` `0; ` `} `

## Python

 `# Python3 program for ` `# the above approach ` ` `  `# Recursive function which will ` `# calculate all the possibilities ` `# recursively ` `def` `recurr(vis, ans): ` `    ``global` `tiles, se ` `    ``if` `(``len``(ans) > ``0``): ` `         `  `        ``# Check that the string ` `        ``# is already there or not ` `        ``if` `(ans ``in` `se): ` `            ``return` `             `  `        ``# Else put in set ` `        ``se[ans] ``=` `1` `         `  `    ``# Run for all the ` `    ``# possibilities ` `    ``for` `i ``in` `range``(``len``(tiles)): ` `         `  `        ``# If already taken ` `        ``# then don't do anything ` `        ``if` `(vis[i]): ` `            ``continue` `        ``vis[i] ``=` `True` `         `  `        ``# Else take it and ` `        ``# call recurr function ` `        ``recurr(vis, ans ``+` `tiles[i]) ` `        ``vis[i] ``=` `False` ` `  `# Driver code ` `tiles ``=` `"AAABBC"` `curr ``=` `"" ` ` `  `se ``=` `dict``() ` `vis ``=` `[``False``] ``*` `(``len``(tiles)) ` `recurr(vis, curr) ` `print``(``len``(se)) ` ` `  `# This code is contributed by mohit kumar 29 `

Output:

```188
```

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Improved By : mohit kumar 29