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.

Source: leetcode | Difficulty: MEDIUM_HARD
Tags: String algorithms

Data structures

Starter code: class Solution:
    def decodeAtIndex(self, S: str, K: int) -> str:
        

Solutions:
class Solution:

	def decodeAtIndex(self, S: str, K: int) -> str:
		size = 0
		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


class Solution:

	def decodeAtIndex(self, S: str, K: int) -> str:
		N = 0
		for (i, c) in enumerate(S):
			N = N * int(c) if c.isdigit() else N + 1
			if K <= N:
				break
		for j in range(i, -1, -1):
			c = S[j]
			if c.isdigit():
				N /= int(c)
				K %= N
			else:
				if K == N or K == 0:
					return c
				N -= 1


class Solution:

	def _decode(self, s, k):
		number_ix = -1
		for i in reversed(range(len(s))):
			if s[i].isnumeric():
				number_ix = i
				break
		if number_ix == -1:
			if k < len(s):
				return (s[k], None)
			else:
				return (None, len(s))
		else:
			(encoded, number, unencoded) = (s[:number_ix], s[number_ix], s[number_ix + 1:])
			(sub_answer, sub_length) = self._decode(encoded, k)
			if sub_answer is not None:
				return (sub_answer, None)
			else:
				if k < sub_length * int(number):
					k = k % sub_length
					(sub_answer, _) = self._decode(encoded, k)
					return (sub_answer, None)
				k = k - sub_length * int(number)
				if k < len(unencoded):
					return (unencoded[k], None)
				else:
					return (None, sub_length * int(number) + len(unencoded))

	def decodeAtIndex(self, S: str, K: int) -> str:
		(answer, _) = self._decode(S, K - 1)
		return answer


class Solution:

	def decodeAtIndex(self, S: str, K: int) -> str:
		fullSize = 0
		for (idx, char) in enumerate(S):
			if char.isdigit():
				fullSize *= int(char)
			else:
				fullSize += 1
			if fullSize > K:
				break
		endIdx = idx
		for idx in range(endIdx, -1, -1):
			if S[idx].isdigit():
				fullSize /= int(S[idx])
				if K > fullSize:
					K %= fullSize
			elif K == 0 or K == fullSize:
				return S[idx]
			else:
				fullSize -= 1