### surajkvm007's blog

By surajkvm007, history, 3 years ago, ,

The following question was asked to my friend in an interview : given a string consisting only of '(' and ')'. find total number of substrings with balanced parenthesis Note: the input string is already balanced.

The only solution i can think of this problem is brute force which takes n^3 time. Is there a faster solution possible.If there is then i would also like to know the build up to that approach.

• +10

 » 3 years ago, # | ← Rev. 3 →   +11 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(NlogN) if you precompute values with a sparse table or O(Nlog2N) if you use segment tree during the binary search.
•  » » 3 years ago, # ^ |   0 raigus in the comment section below gave a linear approach!
 » 3 years ago, # | ← Rev. 5 →   0 Can anybody tell If my approach is right or not ?C++ codestring s;cin>>s;int ans = 0;vector v;for(int i =0;i
•  » » 3 years ago, # ^ |   -9 runs in linear time, Holy sh#t!
 » 3 years ago, # |   0 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; 
 » 3 years ago, # |   +14 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. 
 » 4 weeks ago, # | ← Rev. 3 →   0 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 using namespace std; // #define add accumulate #define ll long long int #define rep(i,k,n) for(int i=k;i>str; ll cnt=0; vector 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<>t; t=1; while(t--) solve(); return 0; }