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);
}
``````