880. Decoded String at Index
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 writtend - 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 digits2
through9
.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.
Solutions
Solution 1: Reverse Thinking
We can first calculate the total length $m$ of the decoded string, then traverse the string from back to front. Each time, we update $k$ to be $k \bmod m$, until $k$ is $0$ and the current character is a letter, then we return the current character. Otherwise, if the current character is a number, we divide $m$ by this number. If the current character is a letter, we subtract $1$ from $m$.
The time complexity is $O(n)$, where $n$ is the length of the string. The space complexity is $O(1)$.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 |
|
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 |
|