1844. Replace All Digits with Characters

You are given a 0-indexed string s that has lowercase English letters in its even indices and digits in its odd indices.

There is a function shift(c, x), where c is a character and x is a digit, that returns the xth character after c.

  • For example, shift('a', 5) = 'f' and shift('x', 0) = 'x'.

For every odd index i, you want to replace the digit s[i] with shift(s[i-1], s[i]).

Return s after replacing all digits. It is guaranteed that shift(s[i-1], s[i]) will never exceed 'z'.

 

Example 1:

Input: s = "a1c1e1"
Output: "abcdef"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('c',1) = 'd'
- s[5] -> shift('e',1) = 'f'

Example 2:

Input: s = "a1b2c3d4e"
Output: "abbdcfdhe"
Explanation: The digits are replaced as follows:
- s[1] -> shift('a',1) = 'b'
- s[3] -> shift('b',2) = 'd'
- s[5] -> shift('c',3) = 'f'
- s[7] -> shift('d',4) = 'h'

 

Constraints:

  • 1 <= s.length <= 100
  • s consists only of lowercase English letters and digits.
  • shift(s[i-1], s[i]) <= 'z' for all odd indices i.

1844. Replace All Digits with Characters
struct Solution;

impl Solution {
    fn replace_digits(s: String) -> String {
        let mut s: Vec<char> = s.chars().collect();
        let n = s.len();
        for i in 0..n {
            if i % 2 == 1 {
                let x = s[i] as u8 - b'0';
                let c = s[i - 1] as u8;
                let d = (c + x) as char;
                s[i] = d;
            }
        }
        s.into_iter().collect()
    }
}

#[test]
fn test() {
    let s = "a1c1e1".to_string();
    let res = "abcdef".to_string();
    assert_eq!(Solution::replace_digits(s), res);
    let s = "a1b2c3d4e".to_string();
    let res = "abbdcfdhe".to_string();
    assert_eq!(Solution::replace_digits(s), res);
}