## The Challenge

An encoded string `S` is given.  To find and write the decoded string to a tape, the encoded string is read one character at a time and the following steps are taken:

• If the character read is a letter, that letter is written onto the tape.
• If the character read is a digit (say `d`), the entire current tape is repeatedly written `d-1` more times in total.

Now for some encoded string `S`, and an index `K`, find and return the `K`-th letter (1 indexed) in the decoded string.

Example 1:

```Input: S = "ha22", K = 5
Output: "h"
Explanation:
The decoded string is "hahahaha".  The 5th letter is "h".
```

Example 2:

```Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation:
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".
```

Note:

1. `2 <= S.length <= 100`
2. `S` will only contain lowercase letters and digits `2` through `9`.
3. `S` starts with a letter.
4. `1 <= K <= 10^9`
5. The decoded string is guaranteed to have less than `2^63` letters.

## The Code

 `````` 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 `````` ``````// Our class class Solution { // The entry method public String decodeAtIndex(String S, int K) { long size = 0; int N = S.length(); // Find size = length of decoded string for (int i = 0; i < N; ++i) { char c = S.charAt(i); if (Character.isDigit(c)) size *= c - '0'; else size++; } for (int i = N-1; i >= 0; --i) { char c = S.charAt(i); K %= size; if (K == 0 && Character.isLetter(c)) return Character.toString(c); if (Character.isDigit(c)) size /= c - '0'; else size--; } throw null; } } ``````