**Characters in Each Bracket Depth - Amazon Top Interview Questions**

### Problem Statement :

You are a given a string s containing "X", "(", and ")". The string has balanced brackets and in between there are some "X"s along with possibly nested brackets recursively. Return the number of "X"s at each depth of brackets in s, from the shallowest depth to the deepest depth. Constraints 2 ≤ n ≤ 100,000 where n is the length of s Example 1 Input s = "(XX(XX(X))X)" Output [3, 2, 1] Explanation There's three "X"s at depth 0. Two "X"s at depth 1. And one "X" at depth 2. Example 2 Input s = "(())" Output [0, 0] Explanation There's no "X"s but depth goes to 1.

### Solution :

` ````
Solution in C++ :
vector<int> solve(string s) {
unordered_map<int, int> m;
int d = 0, maxd = 0;
for (auto &ch : s) {
if (ch == '(')
d += 1;
else if (ch == ')')
d -= 1;
else
m[d] += 1;
maxd = max(maxd, d);
}
vector<int> ret(maxd);
for (auto [dep, nu] : m) {
ret[dep - 1] = nu;
}
return ret;
}
```

` ````
Solution in Java :
import java.util.*;
class Solution {
HashMap<Integer, Integer> hm;
public int[] solve(String s) {
hm = new HashMap();
f(s, 0, -1);
int idx = 0;
int[] res = new int[hm.size() - 1];
for (int i = 0; i < hm.size() - 1; i++) {
res[i] = hm.get(i);
}
return res;
}
public void f(String s, int idx, int depth) {
if (idx >= s.length())
return;
hm.putIfAbsent(depth, 0);
if (s.charAt(idx) == '(') {
f(s, idx + 1, depth + 1);
} else if (s.charAt(idx) == 'X') {
hm.put(depth, hm.get(depth) + 1);
f(s, idx + 1, depth);
} else {
f(s, idx + 1, depth - 1);
}
}
}
```

` ````
Solution in Python :
class Solution:
def solve(self, s):
depth = -1
out = []
for c in s:
if c == "(":
depth += 1
elif c == ")":
depth -= 1
if depth == len(out):
out.append(0)
if c == "X":
out[depth] += 1
return out
```

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