Decoded string at index

/*
 * 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.
 */

impl Solution {
  pub fn decode_at_index(s: String, k: i32) -> String {
    let mut size = 0i64;
    let bytes = s.as_bytes();
    for &c in bytes {
      if c.is_ascii_digit() { size *= (c - b'0') as i64; }
      else { size += 1; }
    }
    let mut k = k as i64;
    for &c in bytes.iter().rev() {
      k %= size;
      if k == 0 && c.is_ascii_alphabetic() { return (c as char).to_string(); }
      if c.is_ascii_digit() { size /= (c - b'0') as i64; }
      else { size -= 1; }
    }
    unreachable!()
  }
}