## Description

https://leetcode.com/problems/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`

.

## Explanation

Find c indexes, and find the minimum distance between each character to c indexes.

## Python Solution

```
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)