"""
880. Decoded String at Index
Medium

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.

"""


class Solution:
    def decodeAtIndex(self, S: str, K: int) -> str:
        sizes = [0]
        N = len(S)
        for c in S:
            if c.isdigit():
                sizes.append(sizes[-1] * int(c))
            else:
                sizes.append(sizes[-1] + 1)
        for i in range(N, 0, -1):
            K %= sizes[i]
            if K == 0 and S[i - 1].isalpha():
                return S[i - 1]
