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.
Input: s = "leet2code3", k = 10 Output: "o" Explanation: The decoded string is "leetleetcodeleetleetcodeleetleetcode". The 10th letter in the string is "o".
Input: s = "ha22", k = 5 Output: "h" Explanation: The decoded string is "hahahaha". The 5th letter is "h".
Input: s = "a2345678999999999999999", k = 1 Output: "a" Explanation: The decoded string is "a" repeated 8301530446056247680 times. The 1st letter is "a".
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.
impl Solution {
pub fn decode_at_index(s: String, k: i32) -> String {
let mut k = k as i64 - 1;
let mut chars = vec![];
let mut length = 0;
for ch in s.bytes() {
chars.push((ch, length));
if ch.is_ascii_lowercase() {
length += 1;
} else {
length *= (ch - b'0') as i64;
}
if length > k {
break;
}
}
while let Some((ch, i)) = chars.pop() {
if ch.is_ascii_lowercase() {
if i == k {
return String::from_utf8(vec![ch]).unwrap();
}
length -= 1;
} else {
length /= (ch - b'0') as i64;
k %= length;
}
}
unreachable!()
}
}