821. Shortest Distance to a Character

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.

---

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        ml=-len(s)
        idx=0
        alist=[]
        for i in range(len(s)):
            if s[i]!=c:
                alist.append(i-ml)
            else:
                alist.append(0)
                ml=i
        print(alist)
        
        ml=len(s)*2
        for i in range(len(s)-1,-1,-1):
            if s[i]!=c:
                alist[i]=min(ml-i,alist[i])
            else:
                ml=i
            
        print(alist)
        return alist