880. Decoded String at Index

Problem Link

Description

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You are given an encoded string s. To decode the 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 d, the entire current tape is repeatedly written d - 1 more times in total.

Given an integer k, return the kth 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 consists of lowercase English letters and digits 2 through 9.
• s starts with a letter.
• 1 <= k <= 109
• It is guaranteed that k is less than or equal to the length of the decoded string.
• The decoded string is guaranteed to have less than 263 letters.

Solution

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Python3

class Solution:
    def decodeAtIndex(self, s: str, k: int) -> str:
        N = len(s)
        curr = 0
 
        for i, x in enumerate(s):
            if x.isdigit():
                curr *= int(x)
            else:
                curr += 1
            
            if curr >= k:
                break
        
        for j in range(i, -1, -1):
            if s[j].isdigit():
                curr //= int(s[j])
                k %= curr
            else:
                if k == 0 or k == curr: return s[j]
 
                curr -= 1