Eye of Sauron Solution

#include <iostream>
#include <string>

using namespace std;

int main()
{
    // input the string
    string str;
    cin>>str;

    // if we've seen an open parentheses
    bool open_eye = false;
    // if we've seen a close parentheses
    bool close_eye = false;
    // number of bars seen before the eye
    int pre_eye = 0;
    // number of bars seen after the eye
    int post_eye = 0;

    // loop through indices
    for(int i=0;i<str.size();i++)
    {
        // if we see a bar
        if(str[i]=='|')
        {
            // if we're looking for a close parentheses, we need to fix it
            if(open_eye)
            {
                cout << "fix\n";
                return 0;
            }
            // increment corresponding operator
            if(!close_eye) pre_eye++;
            else post_eye++;
        }
        // if we see close parentheses
        else if(str[i]==')')
        {
            // if we are not looking to close the eye, we need to fix
            if(!open_eye)
            {
                cout << "fix\n";
                return 0;
            }
            // close the eye
            open_eye = false;
            close_eye = true;
        }
        // if we see an open parentheses
        else
        {
            // if we already saw the eye, we need to fix
            if(open_eye || close_eye)
            {
                cout << "fix\n";
                return 0;
            }
            // open the eye
            open_eye = true;
        }
    }

    // if same number of bars before and after eye, it is correct
    if(pre_eye==post_eye) cout << "correct\n";
    // otherwise it needs to be fixed
    else cout << "fix\n";
}