# Time:  O(n)
# Space: O(1)

# 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".
#
# Note:
#
# 2 <= S.length <= 100
# S will only contain lowercase letters and digits 2 through 9.
# S starts with a letter.
# 1 <= K <= 10^9
# The decoded string is guaranteed to have less than 2^63 letters.

class Solution(object):
    def decodeAtIndex(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        i = 0
        for c in S:
            if c.isdigit():
                i *= int(c)
            else:
                i += 1

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

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

Create decoded-string-at-index.py