Shortest distance to a character

Array Two Pointers String

72.7% acceptance

Feb 22, 2026

3320

193

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.

Solution

Rust

Time O(n)

Space O(n)

LeetCode

solution.rs

/*
 * 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) = 2.
 * 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.
 */

impl Solution {
  pub fn shortest_to_char(s: String, c: char) -> Vec<i32> {
    let bytes = s.as_bytes();
    let cb = c as u8;
    let n = bytes.len();
    let mut res = vec![n as i32; n];
    let mut prev = -(n as i32);
    for i in 0..n {
      if bytes[i] == cb { prev = i as i32; }
      res[i] = i as i32 - prev;
    }
    prev = 2 * n as i32;
    for i in (0..n).rev() {
      if bytes[i] == cb { prev = i as i32; }
      res[i] = res[i].min(prev - i as i32);
    }
    res
  }
}