BINBR - Editorial

editorial
1

PROBLEM LINK:

Practice
Contest

Author: Adarsh Kumar Sinha

DIFFICULTY:

Easy

SOLUTION:

Setter's Solution 
/*----- || Hare Krishna || -----*/
/*  "WHY DO WE FALL, BRUCE?"  */

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

#include<bits/stdc++.h>
#define ll long long
#define endl '\n'
#define elif else if
#define pb push_back
#define pf push_front
#define PI 3.1415926535897932384
#define MOD 1000000007
using namespace std;

char alpha[26] = {'a','b','c','d','e','f','g','h','i','j','k','l','m','n','o','p','q','r','s','t','u','v','w','x','y','z'};

string s;
int t;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
//    freopen("input.txt","r",stdin);
//    freopen("output.txt","w",stdout);
    
    cin >> t;
	while(t--){
		int n;
		cin >> n;
		cin >> s;  // taking input sequence as a string
		
		if(n&1){
			cout << 0 << endl;  // if n is odd, it is not possible
		}
		else{
			
			int open,close;  // number of opening brace, closing brace
			open=close=0;
			
			for(int i=0; i<n; i++){
				
				if(s[i]=='('){
					open++;
				}
				else{
					if(open!=0){
						open--;
					}
					else{
						close++;
					}
				}
				
			}
			
			if((open==0 && close==0) || (open==2 && close==0) || (close==2 && open==0) ) {    // if there was no open or close brace reamining (i.e, each brace got paired)
																							  // ,or if there
																							  // are exactly 2 open brace and 0 close brace, or 2 close brace
																							  // and 0 open brace, only then it is POSSIBLE.
				cout << 1 << endl;
			}
			else{
				cout << 0 << endl;
			}
		}
	}  
}