821. Shortest Distance to a Character

Given a string `s` and a character `c` that occurs in `s`, return an array of integers `answer` where `answer.length == s.length` and `answer[i]` is the distance from index `i` to the closest occurrence of character `c` in `s`.

The distance between two indices `i` and `j` is `abs(i - j)`, where `abs` is the absolute value function.

Example 1:

```Input: s = "loveleetcode", c = "e"
Output: [3,2,1,0,1,0,0,1,2,2,1,0]
Explanation: The character 'e' appears at indices 3, 5, 6, and 11 (0-indexed).
The closest occurrence of 'e' for index 0 is at index 3, so the distance is abs(0 - 3) = 3.
The closest occurrence of 'e' for index 1 is at index 3, so the distance is abs(1 - 3) = 3.
For index 4, there is a tie between the 'e' at index 3 and the 'e' at index 5, but the distance is still the same: abs(4 - 3) == abs(4 - 5) = 1.
The closest occurrence of 'e' for index 8 is at index 6, so the distance is abs(8 - 6) = 2.
```

Example 2:

```Input: s = "aaab", c = "b"
Output: [3,2,1,0]
```

Constraints:

• `1 <= s.length <= 104`
• `s[i]` and `c` are lowercase English letters.
• It is guaranteed that `c` occurs at least once in `s`.

Rust Solution

``````struct Solution;

impl Solution {
fn shortest_to_char(s: String, c: char) -> Vec<i32> {
let s: Vec<char> = s.chars().collect();
let mut prev: Option<usize> = None;
let n = s.len();
let mut res: Vec<i32> = vec![n as i32; n];
for i in 0..n {
if s[i] == c {
prev = Some(i);
}
if let Some(j) = prev {
res[i] = i32::min(res[i], (i - j) as i32);
}
}
prev = None;
for i in (0..n).rev() {
if s[i] == c {
prev = Some(i);
}
if let Some(j) = prev {
res[i] = i32::min(res[i], (j - i) as i32);
}
}
res
}
}

#[test]
fn test() {
let s = "loveleetcode".to_string();
let t = vec![3, 2, 1, 0, 1, 0, 0, 1, 2, 2, 1, 0];
assert_eq!(Solution::shortest_to_char(s, 'e'), t);
}
``````

Having problems with this solution? Click here to submit an issue on github.