880. Decoded String at Index

An encoded string S is given.  To find and write the decoded 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 (say d), the entire current tape is repeatedly written d-1 more times in total.

Now for some encoded string S, and an index K, find and return the K-th 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 will only contain lowercase letters and digits 2 through 9.
  • S starts with a letter.
  • 1 <= K <= 10^9
  • It's guaranteed that K is less than or equal to the length of the decoded string.
  • The decoded string is guaranteed to have less than 2^63 letters.

880. Decoded String at Index
struct Solution;

impl Solution {
    fn decode_at_index(s: String, k: i32) -> String {
        let mut k = k as u64;
        let s: Vec<char> = s.chars().collect();
        let n = s.len();
        let mut size = 0;
        for i in 0..n {
            match s[i] {
                '0'..='9' => {
                    size *= (s[i] as u8 - b'0') as u64;
                }
                _ => {
                    size += 1;
                }
            }
        }
        let mut res = "".to_string();
        for i in (0..n).rev() {
            k %= size;
            match s[i] {
                '0'..='9' => {
                    size /= (s[i] as u8 - b'0') as u64;
                }
                _ => {
                    if k == 0 {
                        res = s[i].to_string();
                        break;
                    } else {
                        size -= 1;
                    }
                }
            }
        }
        res
    }
}

#[test]
fn test() {
    let s = "leet2code3".to_string();
    let k = 10;
    let res = "o".to_string();
    assert_eq!(Solution::decode_at_index(s, k), res);
}