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) = 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.

Example 2:
Input: s = "aaab", c = "b"
Output: [3,2,1,0]

Constraints:

• 1 <= s.length <= 10^4
• s[i] and c are lowercase English letters.
• It is guaranteed that c occurs at least once in s.

Solution in python:

class Solution:
    def shortestToChar(self, s: str, c: str) -> List[int]:
        index = []
        result = [0 for i in range(len(s))]
        for i in range(len(s)):
            if s[i] == c:
                index.append(i)
        i = 0
        while i < index[0]:
            result[i] = abs(i-index[0])
            i += 1

        for i in range(len(index)-1):
            print(result)
            left = index[i]
            right = index[i+1]
            mid, r = divmod((left+right), 2)
            j = left
            while j <= mid:
                result[j] = (j-left)
                j += 1
            k = mid
            if r == 1:
                k = mid + 1
            while k < right:
                result[k] = (right-k)
                k += 1

        i = index[-1]
        while i < len(s):
            result[i] = abs(i-index[-1])
            i += 1
        return result