# LeetCode 821. Shortest Distance to a Character

## 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 <= 104`
• `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)