# Shortest Distance To A Character Problem

## Description

LeetCode Problem 821.

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:

``````1
2
3
4
5
6
7
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:

``````1
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.

## Sample C++ Code

``````1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
class Solution {
public:
vector<int> shortestToChar(string s, char c) {
vector<int> res;
int prev_char = -s.size();
for (int i = 0; i < s.size(); i++) {
if (s[i] == c)
prev_char = i;
res.push_back(i - prev_char);
}

for (int i = prev_char; i >= 0; i--) {
if (s[i] == c)
prev_char = i;
res[i] = min(res[i], prev_char - i);
}
return res;
}
};
``````