## 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, S, ..., S[i].

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

Complexity Analysis

• Time Complexity: , where is the length of S.

• Space Complexity: .

Analysis written by: @awice.