Identify some positions in the parenthesis sequence

link ：C. Recover an RBS

The question ：
Given the bracket sequence , Some of these locations are unknown , If these unknown locations can determine what to fill , It would be YES, otherwise NO.

practice ：
cnt Statistical question mark ？ The number of ,sum Is the number of left parentheses - Number of right parentheses . When sum + cnt == 1 when , It can be determined that cnt individual ？ All for （. At this time, you just need to empty cnt, And then let sum = 1.
Last , If cnt == sum, Explain the final remaining cnt individual ？ It has to be for ）, That is, you can output YES 了 .

Code ：

void solve()
{
    
    string s;
    cin >> s;
    int cnt = 0, sum = 0;
    for (auto i : s)
    {
    
        if (i == '(') sum++;
        else if (i == ')') sum--;
        else cnt++;
        if (cnt + sum == 1)
        {
    
            cnt = 0;
            sum = 1;
        }
    }
    if (sum == cnt) cout << "YES\n";
    else cout << "NO\n";
}