## 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#

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