821 - Shortest Distance to a Character\.

Easy

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:

• <code>1 <= s.length <= 10⁴</code>

• s[i] and c are lowercase English letters.

• It is guaranteed that c occurs at least once in s.