# HackerEarth Bracket sequences problem solution

In this HackerEarth Bracket sequences problem solution A bracket sequence is a string that contains only characters '(' and ')'.

A correct bracket sequence is a bracket sequence that can be transformed into a correct arithmetic expression by inserting characters '1' and '+' between the original characters of the sequence. For example, bracket sequences '()()' and '(())' are correct. The resulting expressions of these sequences are: '(1)+(1)' and '((1+1)+1)'. However, '(', ')(', and '(' are incorrect bracket sequences.

You are given a bracket sequence s(s1 s2...sn), where si denotes the type of 's bracket (open or close). It is not mandatory that s is necessarily correct. Your task is to determine the number of i's such that sisi+1...sns1s2...si-1 is a correct bracket sequence.

## HackerEarth Bracket sequences problem solution.

`#include <bits/stdc++.h> #define mp make_pair#define pb push_back#define f first#define s second#define ll long long#define forn(i, a, b) for(int i = (a); i <= (b); ++i)#define forev(i, b, a) for(int i = (b); i >= (a); --i)#define VAR(v, i) __typeof( i) v=(i)#define forit(i, c) for(VAR(i, (c).begin()); i != (c).end(); ++i)#define all(x) (x).begin(), (x).end()#define sz(x) ((int)(x).size())#define file(s) freopen(s".in","r",stdin); freopen(s".out","w",stdout); using namespace std; const int maxn = (int)1e6 + 100;const int mod = (int)1e9 + 7;const int P = (int) 1e6 + 7; const double pi = acos(-1.0);#define inf mod typedef long double ld;typedef pair<int, int> pii;typedef pair<ll, ll> pll;typedef vector<int> vi;   typedef vector<ll> Vll;               typedef vector<pair<int, int> > vpii;typedef vector<pair<ll, ll> > vpll;                        char s[maxn];int n, cnt, bal, mn = inf;void solve(){    scanf("%s", s + 1);    n = strlen(s + 1);    forn(i, 1, n){        if(s[i] == '(') bal++;        else bal--;        if(mn > bal) mn = bal, cnt = 0;        if(mn == bal) cnt++;    }    if(bal){        puts("0");        return;    }    printf("%d", cnt);}main () {    int t = 1;    while(t--)        solve(); }       `

### Second solution

`#include <bits/stdc++.h>using namespace std;typedef long long ll; const int maxn = 3e2 + 14;string s;int main(){    ios::sync_with_stdio(0), cin.tie(0);    cin >> s;    map<int, int> mp;    int su = 0;    for(auto c : s){        su += (c == '(' ? +1 : -1);        mp[su]++;    }    cout << (su == 0) * mp.begin() -> second << '\n';}`

1. it would be very helpful if you right some logic for the for loop
what exactly you are doing here??

forn(i, 1, n){
if(s[i] == '(') bal++;
else bal--;
if(mn > bal) mn = bal, cnt = 0;
if(mn == bal) cnt++;
}

2. please help me, how should i know for a given bracket sequence there exist a valid sequence and how would i found that valid sequence??

like for
1) s = ")()()(" here valid sequence exist and the valid sequence is ()()()
2) s = "(()))" here valid sequence does not exist