Package g0801_0900.s0821_shortest_distance_to_a_character

Class Solution

  • public class Solution
extends Object
    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:

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

    • Constructor Detail

      • Solution

        public Solution()

    • Method Detail

      • shortestToChar

        public int[] shortestToChar​(String s,
                            char c)