human
I found an interesting problem on https://leetcode.com/problems/decoded-string-at-index/:
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 = "leet2code3", K = 10
Output: "o"
Explanation: 
The decoded string is "leetleetcodeleetleetcodeleetleetcode".
The 10th letter in the string is "o".

Example 2:
Input: S = "ha22", K = 5
Output: "h"
Explanation: 
The decoded string is "hahahaha".  The 5th letter is "h".

Example 3:
Input: S = "a2345678999999999999999", K = 1
Output: "a"
Explanation: 
The decoded string is "a" repeated 8301530446056247680 times.  The 1st letter is "a".



 
Constraints:

2 <= S.length <= 100
S will only contain lowercase letters and digits 2 through 9.
S starts with a letter.
1 <= K <= 10^9
It's guaranteed that K is less than or equal to the length of the decoded string.
The decoded string is guaranteed to have less than 2^63 letters.

I tried it in Python, but could not do it. Can you solve it?
assistant
#Here is Python code 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:
.

class Solution:
    def decodeAtIndex(self, S: str, K: int) -> str:
        size = 0
        # Find size = length of decoded string
        for c in S:
            if c.isdigit():
                size *= int(c)
            else:
                size += 1

        for c in reversed(S):
            K %= size
            if K == 0 and c.isalpha():
                return c

            if c.isdigit():
                size /= int(c)
            else:
                size -= 1