Given a string
s and a character
c that occurs in
s, return an array of integers
answer.length == s.length and
answer[i] is the distance from index
i to the closest occurrence of character
The distance between two indices
abs(i - j), where
abs is the absolute value function.
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.
Input: s = "aaab", c = "b" Output: [3,2,1,0]
1 <= s.length <= 104
care lowercase English letters.
- It is guaranteed that
coccurs at least once in
Find c indexes, and find the minimum distance between each character to c indexes.
class Solution: def shortestToChar(self, s: str, c: str) -> List[int]: c_indexes =  for i, s_c in enumerate(s): if s_c == c: c_indexes.append(i) results =  for i, s_c in enumerate(s): min_distance = sys.maxsize for c_i in c_indexes: min_distance = min(abs(i - c_i), min_distance) results.append(min_distance) return results
- Time Complexity: O(N^2)
- Space Complexity: O(N)