1. Abridged Problem Statement
Given a string s (length up to 10^5) consisting of letters, spaces, parentheses '(' and ')', and a few punctuation marks, determine how many of the round brackets are parts of emoticons.
Definition:
- A bracket is normal (i.e. not part of an emoticon) if:
  - It's a '(' and its nearest round bracket to the right is a ')', and all characters strictly between them are either letters or spaces.
  - Or symmetrically: it's a ')' and its nearest round bracket to the left is a '(', and all characters strictly between are letters or spaces.
- All other parentheses are parts of emoticons.
Output the total count of emoticon parentheses.

2. Detailed Editorial

Problem restated:
We need to classify each '(' or ')' in the string as either "normal" (a valid parenthesis pair enclosing only letters/spaces) or "emoticon" and count the emoticon ones.

Key observations and steps:

1. We must quickly find, for every position i,
   - the nearest bracket to the left (prev_bracket[i]),
   - the nearest bracket to the right (next_bracket[i]).
   We can fill these arrays in one left-to-right and one right-to-left pass in O(n).

2. We need to check, for a candidate pair (i, j), that there are no "forbidden" characters between i and j.
   Forbidden characters are punctuation marks other than letters and spaces, i.e. anything other than isalpha(c) or ' ' or parentheses themselves.
   To answer "are there any forbidden chars in s[i+1..j-1]?" in O(1), build a prefix-sum array cnt_non_letters[k] = count of forbidden chars up to index k.

3. Counting emoticon brackets:
   - First assume all parentheses are emoticon: ans = total '(' and ')'.
   - Then detect each "normal" bracket and decrement the count for it.
     - For each i with s[i]=='(':
       let j = next_bracket[i]. If j exists, and prev_bracket[j]==i (so they are each other's nearest), and there are no forbidden chars between i and j, then this '(' is normal-subtract 1.
     - Similarly for s[i]==')', using prev_bracket[i] and checking next_bracket.

Note: the code tracks prev_bracket only for '(' (not ')') and next_bracket only for ')' (not '('), which is sufficient for the normal-pair check since a normal pair consists of '(' followed immediately (in bracket order) by ')'.

Complexities:
- Time: O(n) for building prev_bracket, next_bracket, prefix sums, and one more O(n) scan.
- Memory: O(n).

Example "Hi:) (it is me) I have bad news:-((" :
- Brackets around "it is me" are normal, so we do not count them.
- All others (':)', ':-((' ) are emoticon brackets -> total 3.

3. C++ Solution

```cpp
#include <bits/stdc++.h>

using namespace std;

template<typename T1, typename T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& x) {
    return out << x.first << ' ' << x.second;
}

template<typename T1, typename T2>
istream& operator>>(istream& in, pair<T1, T2>& x) {
    return in >> x.first >> x.second;
}

template<typename T>
istream& operator>>(istream& in, vector<T>& a) {
    for(auto& x: a) {
        in >> x;
    }
    return in;
};

template<typename T>
ostream& operator<<(ostream& out, const vector<T>& a) {
    for(auto x: a) {
        out << x << ' ';
    }
    return out;
};

string s;

void read() { getline(cin, s); }

void solve() {
    // A bracket pair is a "real" pair (not part of an emoticon) when an '(' has
    // its nearest bracket to the right being a ')', that ')' has this '(' as its
    // nearest bracket to the left, and everything between them is only spaces
    // and Latin letters (and the pair is not the empty "()"). All other
    // brackets are emoticon parts, which is what we count.
    //
    // prev_bracket[i] / next_bracket[i] give, for position i, the index of the
    // nearest '(' to the left / ')' to the right (or itself if i is that
    // bracket). cnt_non_letters is a prefix sum of characters that are neither
    // letters, spaces, nor brackets, so get_non_letters(l, r) tests emptiness
    // of the inside. We start by counting every bracket, then for each matched,
    // clean pair we subtract its two brackets back out.

    int n = s.size();
    vector<int> prev_bracket(n, -1);
    vector<int> next_bracket(n, -1);
    vector<int> cnt_non_letters(n, 0);
    for(int i = 0; i < n; i++) {
        if(s[i] == '(') {
            prev_bracket[i] = i;
        } else {
            prev_bracket[i] = i ? prev_bracket[i - 1] : -1;
        }

        if(!isalpha(s[i]) && s[i] != ' ' && s[i] != '(' && s[i] != ')') {
            cnt_non_letters[i]++;
        }
        cnt_non_letters[i] += i ? cnt_non_letters[i - 1] : 0;
    }

    auto get_non_letters = [&](int l, int r) {
        return cnt_non_letters[r] - (l ? cnt_non_letters[l - 1] : 0);
    };

    for(int i = n - 1; i >= 0; i--) {
        if(s[i] == ')') {
            next_bracket[i] = i;
        } else {
            next_bracket[i] = i < n - 1 ? next_bracket[i + 1] : -1;
        }
    }

    int ans = 0;
    for(int i = 0; i < n; i++) {
        if(s[i] == '(' || s[i] == ')') {
            ans++;
        }

        if(s[i] == '(' && next_bracket[i] != -1 &&
           prev_bracket[next_bracket[i]] == i &&
           get_non_letters(i, next_bracket[i]) == 0) {
            ans--;
        }

        if(s[i] == ')' && prev_bracket[i] != -1 &&
           next_bracket[prev_bracket[i]] == i &&
           get_non_letters(prev_bracket[i], i) == 0) {
            ans--;
        }
    }

    cout << ans << '\n';
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    int T = 1;
    // cin >> T;
    for(int test = 1; test <= T; test++) {
        read();
        // cout << "Case #" << test << ": ";
        solve();
    }

    return 0;
}
```

4. Python Solution

```python
import sys

def count_emoticons(s: str) -> int:
    n = len(s)
    # Arrays to store nearest bracket indices
    prev_b = [-1] * n
    next_b = [-1] * n
    # Prefix sum of forbidden characters
    bad = [0] * n

    # Build prev_b and bad
    for i, ch in enumerate(s):
        # prev bracket
        if ch == '(' or ch == ')':
            prev_b[i] = i
        else:
            prev_b[i] = prev_b[i-1] if i > 0 else -1
        # forbidden if not letter, not space, not bracket
        is_forbidden = not (ch.isalpha() or ch == ' ' or ch in '()')
        bad[i] = bad[i-1] + (1 if is_forbidden else 0)

    # Build next_b in reverse
    for i in range(n-1, -1, -1):
        if s[i] == '(' or s[i] == ')':
            next_b[i] = i
        else:
            next_b[i] = next_b[i+1] if i+1 < n else -1

    # Helper to count forbidden chars in s[l..r]
    def count_bad(l, r):
        if l > r:
            return 0
        return bad[r] - (bad[l-1] if l > 0 else 0)

    # Start by counting all parentheses as emoticons
    ans = sum(1 for ch in s if ch in '()')

    # Subtract normal '('
    for i, ch in enumerate(s):
        if ch == '(':
            j = next_b[i]
            if j != -1 and s[j] == ')' and prev_b[j] == i and count_bad(i+1, j-1) == 0:
                ans -= 1
        elif ch == ')':
            j = prev_b[i]
            if j != -1 and s[j] == '(' and next_b[j] == i and count_bad(j+1, i-1) == 0:
                ans -= 1

    return ans

if __name__ == "__main__":
    s = sys.stdin.readline().rstrip('\n')
    print(count_emoticons(s))
```

5. Compressed Editorial
- Precompute nearest '(' to the left (prev_bracket) and nearest ')' to the right (next_bracket) via single passes.
- Build a prefix sum of characters that are neither letters, spaces, nor brackets.
- Initially count all '(' and ')' as emoticons.
- For each '(' at i: if its nearest right bracket j is a ')' and prev_bracket[j]==i and no forbidden chars lie between i and j, subtract 1 (it is a normal pair).
- Apply the symmetric check for each ')'.
- Print the remaining count.
