### Problem Statement :

```Given a string s and a character c, return a new list of integers of the same length as s where for each index i its value is set the closest distance of s[i] to c. You can assume c exists in s.

Constraints

n ≤ 100,000 where n is the length of s

Example 1

Input

s = "aabaab"

c = "b"

Output

[2, 1, 0, 1, 1, 0]```

### Solution :

```                        ```Solution in C++ :

vector<int> solve(string s, string c) {
vector<int> res(s.size(), s.size());
set<int> pos;

for (int i = 0; i < s.size(); i++)
if (s[i] == c) pos.insert(i);

for (int i = 0; i < s.size(); i++) {
auto it = pos.lower_bound(i);
// check your right
if (it != pos.end()) res[i] = *it - i;
// check your left
if (it != pos.begin()) res[i] = min(res[i], i - *prev(it));
}
return res;
}```
```

```                        ```Solution in Java :

import java.util.*;

class Solution {
public int[] solve(String s, String c) {
Integer[] ret = new Integer[s.length()];
Queue<Integer> q = new LinkedList();
for (int i = 0; i < s.length(); i++) {
if (String.valueOf(s.charAt(i)).equals(c)) {
q.offer(i);
}
}
Set<Integer> vis = new HashSet();
int dist = 0;
while (!q.isEmpty()) {
int size = q.size();
for (int x = 0; x < size; x++) {
int temp = q.poll();
if (temp < 0 || temp >= s.length() || vis.contains(temp) || ret[temp] != null) {
continue;
}
ret[temp] = dist;
q.offer(temp + 1);
q.offer(temp - 1);
}
dist++;
}
int[] convert = new int[ret.length];
for (int i = 0; i < ret.length; i++) convert[i] = ret[i];
return convert;
}
}```
```

```                        ```Solution in Python :

class Solution:
def solve(self, s, c):
result = [inf] * len(s)

last_found = inf

for pos, char in enumerate(s):
if char == c:
result[pos] = 0
last_found = 0
elif last_found != inf:
last_found += 1
result[pos] = last_found

for pos in range(len(s) - 2, -1, -1):
if result[pos + 1] + 1 < result[pos]:
result[pos] = result[pos + 1] + 1

return result```
```

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