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Problem

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]

Code

class Solution {
    public int[] shortestToChar(String s, char c) {
        int len = s.length();
        int[] ans = new int[len];
        int prev = Integer.MIN_VALUE / 2;

        for (int i = 0; i < len; ++i) {
            if (s.charAt(i) == c) prev = i;
            ans[i] = i - prev;
        }

        prev = Integer.MAX_VALUE / 2;
        for (int i = len - 1; i >= 0; --i) {
            if (s.charAt(i) == c) prev = i;
            ans[i] = Math.min(ans[i], prev - i);
        }

        return ans;
    }
}