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.

Difficulty: interview

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
        # 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


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

'''
this problem is recursive
Split the string S into (encoded substring, number, unencoded string)
If K is in the unencoded part, then we can just return the correct value
If K is in the encoded part, we recurse

I either need the length of the unencoded string, or try the encoded part first and
return (None, length) if it isn't there

'''
class Solution:
    # returns (answer, none) or (none, length of unencoded string)
    # k is 0-indexed
    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:
            # string is all unencoded
            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
            else:
                if K==0 or K == fullSize:
                    return S[idx]
                else:
                    fullSize -=1

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
                


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):
            K%=fullSize
            if K==0 and S[idx].isalpha():
                return S[idx]
            if S[idx].isdigit():
                fullSize /= int(S[idx])
            else:
                fullSize -=1

class Solution:
    def decodeAtIndex(self, S, K):
        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