Shortest Distance to a Character

Problem statement

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.

My solution

/**
 * @param {string} s
 * @param {character} c
 * @return {number[]}
 */
var shortestToChar = function(s, c) {
    let firstIdx = s.indexOf(c)
    const answers = Array.from({
        length: s.length
    }, () => 0)
    for (let i = 0; i < s.length; i++) {
        const char = s.charAt(i)
        
        if (char === c) {
            firstIdx = i;
        } else {
            answers[i] = Math.abs(i - firstIdx)
        }
    }
    
    let lastIdx = s.lastIndexOf(c)
    
    for (let i = s.length - 1; i >= 0; i--) {
        const char = s.charAt(i)
        if (char === c) {
            lastIdx = i
        } else {
            answers[i] = Math.min(answers[i], Math.abs(i - lastIdx))
        }
    }
    
    return answers
};