880. Decoded String at Index

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.

Solution

Approach 1: Work Backwards

Intuition

If we have a decoded string like appleappleappleappleappleapple and an index like K = 24, the answer is the same if K = 4.

In general, when a decoded string is equal to some word with size length repeated some number of times (such as apple with size = 5 repeated 6 times), the answer is the same for the index K as it is for the index K % size.

We can use this insight by working backwards, keeping track of the size of the decoded string. Whenever the decoded string would equal some word repeated d times, we can reduce K to K % (word.length).

Algorithm

First, find the length of the decoded string. After, we'll work backwards, keeping track of size: the length of the decoded string after parsing symbols S[0], S[1], ..., S[i].

If we see a digit S[i], it means the size of the decoded string after parsing S[0], S[1], ..., S[i-1] will be size / Integer(S[i]). Otherwise, it will be size - 1.

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

Complexity Analysis