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) = 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.
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
class Solution: def shortestToChar(self, S: str, C: str) -> List[int]: res =  pos_c =  for i in range(len(S)): if S[i]==C: pos_c.append(i) for i in range(len(S)): m=99999999 for j in pos_c: m = min(m,abs(i-j)) res.append(m) return (res)