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880. Decoded String at Index

Difficulty Topics

Description

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

decoded-string-at-index.py
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