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 `bad[k]` = count of forbidden chars up to index k.  
   Then the count in [l..r] is `bad[r] - bad[l-1]`.

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 `i ← prev_bracket[i]` and `next_bracket`.

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 with Detailed Comments  
```cpp
#include <bits/stdc++.h>
using namespace std;

// Overload << and >> for convenience with pairs and vectors
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;

// Read the entire input line into s
void read() {
    getline(cin, s);
}

void solve() {
    int n = s.size();

    // prev_bracket[i] = index of the last '(' or ')' at or before i
    vector<int> prev_bracket(n, -1);
    // next_bracket[i] = index of the next '(' or ')' at or after i
    vector<int> next_bracket(n, -1);
    // cnt_non_letters[i] = number of forbidden chars (punctuation other than spaces/letters/parentheses)
    // up to and including i
    vector<int> cnt_non_letters(n, 0);

    // Build prev_bracket and cnt_non_letters in one pass
    for(int i = 0; i < n; i++) {
        // Track the last bracket
        if (s[i] == '(' || s[i] == ')') {
            prev_bracket[i] = i;
        } else {
            prev_bracket[i] = (i > 0 ? prev_bracket[i - 1] : -1);
        }
        // Count forbidden characters
        bool forbidden = true;
        if (isalpha(s[i]) || s[i] == ' ' || s[i] == '(' || s[i] == ')')
            forbidden = false;
        cnt_non_letters[i] = (i > 0 ? cnt_non_letters[i - 1] : 0) + (forbidden ? 1 : 0);
    }

    // Helper lambda to get number of forbidden chars in [l..r]
    auto get_bad = [&](int l, int r) {
        if (l > r) return 0;
        return cnt_non_letters[r] - (l > 0 ? cnt_non_letters[l - 1] : 0);
    };

    // Build next_bracket in a reverse pass
    for(int i = n - 1; i >= 0; i--) {
        if (s[i] == '(' || s[i] == ')') {
            next_bracket[i] = i;
        } else {
            next_bracket[i] = (i + 1 < n ? next_bracket[i + 1] : -1);
        }
    }

    // Start by counting all parentheses as emoticons
    int ans = 0;
    for(char c: s) {
        if (c == '(' || c == ')') ans++;
    }

    // Now subtract 1 for each normal '(' and each normal ')'
    for(int i = 0; i < n; i++) {
        if (s[i] == '(') {
            int j = next_bracket[i];
            // Check if j exists, is a ')', and they are each other's nearest bracket
            if (j != -1 && s[j] == ')' &&
                prev_bracket[j] == i &&
                // and ensure no forbidden chars between i+1..j-1
                get_bad(i + 1, j - 1) == 0) {
                ans--;
            }
        }
        else if (s[i] == ')') {
            int j = prev_bracket[i];
            // Check if j exists, is a '(', and they are each other's nearest bracket
            if (j != -1 && s[j] == '(' &&
                next_bracket[j] == i &&
                // and ensure no forbidden chars between j+1..i-1
                get_bad(j + 1, i - 1) == 0) {
                ans--;
            }
        }
    }

    cout << ans << "\n";
}

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

    // Read one line and solve
    read();
    solve();
    return 0;
}
```

4. Python Solution with Detailed Comments  
```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 bracket to left/right and a prefix sum of forbidden chars.  
- Initially count all '(' and ')' as emoticons.  
- For each bracket, if it pairs with its nearest opposite bracket and encloses only spaces/letters, it’s **normal**—subtract 1.  
- Print the remaining count.