# [Solved] 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.

## Question

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`.

## Python Solution

```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)```