Consider open parenthesis as 1 and closing parenthesis as -1, so necessary conditions for substring to be balanced are:

• Sum of all values is 0.
• There isn't a prefix with negative sum.

OK, now compute L[i] = total sum of elements from index 1 to i. So for each left index l that is an open parenthesis (value 1), find with binary search maximum possible right index r such that there isn't an index i between l and r with L[i] < L[l] - 1. This ensures condition 2. You can do this step with a segment tree or a sparse table. Now to satisfy condition 1, find all indexes i between l and r such that L[i] = L[l] - 1. You can do this one with binary search too, for example by storing positions for given sum s in a vector Pos[s].

Overall complexity of the algorithm is O(N_logN) if you precompute values with a sparse table or O(N_log²N) if you use segment tree during the binary search.

Can anybody tell If my approach is right or not ?

C++ code

string s;

cin>>s;

int ans = 0;

vector v;

for(int i =0;i<s.length();i++){

if(s[i]=='('){

v.push_back(1);

}

else{

int count = 0;

 while(v.top()!=1){

     v.pop_back();

     count++;

  }

v.pop_back();

ans += count*(count+1)/2;

v.push_back(0);

}

}

int n =v.size();

ans += n*(n+1)/2;

v.clear();

cout<<ans<<endl;

Time complexity will be O(N)

if the length of string is not large you can use a simple O(n^2) solution .

C++ code : 
        string s;
	cin >> s;
	int result = 0;
	for (int i = 0; i < s.size(); ++i){
		int counter = 0;
		for (int j = i;counter >= 0 &&  j < s.size(); ++j){
			if (s[j] == '(')counter++;
			else counter--;
			if (counter == 0) result++;
		}
	}
	cout << result << endl;

I think there is an O(n) solution.
First you can find all the opening parenthesis for closing ones using stack in O(n).
Use array d[n+1] for keeping the count initialized with 0. 
Then for every opening closing pair with opening at i and closing at j set 
d[j] = 1 + d[i-1]. This can be done for all indices from 1 to n in O(n). 
Finally for answer accumulate all d[i]'s.

For example s =    ( ) ( ( ) ) ( ) ( ( ) ( ) ). 
            d = {0,0,1,0,0,1,2,0,3,0,0,1,0,2,4};

So, the answer will be 14.

We can do this , lets define the level of any "()" to be the number of opening brackets under which it appears so just count number of "()" and for every "()" also add in the answer the number of balanced substrings appearing in that level .

#/*
	AUTHOR:shivam51
	IIESTS
*/
  #include<bits/stdc++.h>
  using namespace std;
  //
  #define add accumulate
  #define ll long long int
  #define rep(i,k,n) for(int i=k;i<n;i++)
  //
  //
	void solve(){
		string str;
		cin>>str;
		ll cnt=0;
		vector<char> vec;
		vll arr(str.length()/2+1,0);
		int pre=0;
		rep(i,0,str.length()){
			if(str[i]=='('){
				++pre;
				vec.pb('(');
			}
			else{
				++cnt;
				vec.pop_back();
				cnt+=arr[pre];
				arr[pre]++;
				--pre;
			}
		}
		cout<<cnt;
	}
  int main(){ 
	hs;
	ll t;
	// cin>>t;
	t=1;
	while(t--)
	    solve();
	return 0; 
}