# Range Queries for Longest Correct Bracket Subsequence Set | 2

Given a bracket sequence or in other words a string S of length n, consisting of characters ‘(‘ and ‘)’. Find the length of the maximum correct bracket subsequence of sequence for a given query range. Note: A correct bracket sequence is the one that has matched bracket pairs or which contains another nested correct bracket sequence. For e.g (), (()), ()() are some correct bracket sequence.

Examples:

```Input : S = ())(())(())(
Start Index of Range = 0,
End Index of Range = 11
Output : 10
Explanation:  Longest Correct Bracket Subsequence is ()(())(())

Input : S = ())(())(())(
Start Index of Range = 1,
End Index of Range = 2
Output : 0
```

Approach : In the Previous post (SET 1) we discussed a solution that works in O(long) for each query, now is this post we will going to see a solution that works in O(1) for each query.
Idea is based on the Post length of the longest valid balanced substring If we marked indexes of all Balanced parentheses/bracket in a temporary array (here we named it BCP[], BOP[] ) then we answer each query in O(1) time.
Algorithm :

```stack is used to get the index of balance bracket.
Travese a string from 0 ..to n
IF we seen a closing bracket,
( i.e., str[i] = ')' && stack is not empty )

Then mark both "open & close" bracket indexes as 1.
BCP[i] = 1;
BOP[stk.top()] = 1;

And At last, stored cumulative sum of BCP[] & BOP[]
Run a loop from 1 to n
BOP[i] +=BOP[i-1], BCP[i] +=BCP[i-1]
```

Now you can answer each query in O(1) time

```(BCP[e] - BOP[s-1]])*2;
```

Below is the implementation of above idea.

 `// CPP code to answer the query in constant time ` `#include ` `using` `namespace` `std; ` ` `  `/* ` `BOP[] stands for "Balanced open parentheses"  ` `BCP[] stands for "Balanced close parentheses" ` ` `  `*/` ` `  `// function for precomputation ` `void` `constructBlanceArray(``int` `BOP[], ``int` `BCP[], ` `                          ``char``* str, ``int` `n) ` `{ ` ` `  `    ``// Create a stack and push -1 as initial index to it. ` `    ``stack<``int``> stk; ` ` `  `    ``// Initialize result ` `    ``int` `result = 0; ` ` `  `    ``// Traverse all characters of given string ` `    ``for` `(``int` `i = 0; i < n; i++) { ` `        ``// If opening bracket, push index of it ` `        ``if` `(str[i] == ``'('``) ` `            ``stk.push(i); ` ` `  `        ``else` `// If closing bracket, i.e., str[i] = ')' ` `        ``{ ` `            ``// If closing bracket, i.e., str[i] = ')' ` `            ``// && stack is not empty then mark both ` `            ``// "open & close" bracket indexs as 1 . ` `            ``// Pop the previous opening bracket's index ` `            ``if` `(!stk.empty()) { ` `                ``BCP[i] = 1; ` `                ``BOP[stk.top()] = 1; ` `                ``stk.pop(); ` `            ``} ` ` `  `            ``// If stack is empty. ` `            ``else` `                ``BCP[i] = 0; ` `        ``} ` `    ``} ` ` `  `    ``for` `(``int` `i = 1; i < n; i++) { ` `        ``BCP[i] += BCP[i - 1]; ` `        ``BOP[i] += BOP[i - 1]; ` `    ``} ` `} ` ` `  `// Function return output of each query in O(1) ` `int` `query(``int` `BOP[], ``int` `BCP[], ` `          ``int` `s, ``int` `e) ` `{ ` `    ``if` `(BOP[s - 1] == BOP[s]) { ` `        ``return` `(BCP[e] - BOP[s]) * 2; ` `    ``} ` ` `  `    ``else` `{ ` `        ``return` `(BCP[e] - BOP[s] + 1) * 2; ` `    ``} ` `} ` ` `  `// Driver program to test above function ` `int` `main() ` `{ ` ` `  `    ``char` `str[] = ``"())(())(())("``; ` `    ``int` `n = ``strlen``(str); ` ` `  `    ``int` `BCP[n + 1] = { 0 }; ` `    ``int` `BOP[n + 1] = { 0 }; ` ` `  `    ``constructBlanceArray(BOP, BCP, str, n); ` ` `  `    ``int` `startIndex = 5, endIndex = 11; ` ` `  `    ``cout << ``"Maximum Length Correct Bracket"` `            ``" Subsequence between "` `         ``<< startIndex << ``" and "` `<< endIndex << ``" = "` `         ``<< query(BOP, BCP, startIndex, endIndex) << endl; ` ` `  `    ``startIndex = 4, endIndex = 5; ` `    ``cout << ``"Maximum Length Correct Bracket"` `            ``" Subsequence between "` `         ``<< startIndex << ``" and "` `<< endIndex << ``" = "` `         ``<< query(BOP, BCP, startIndex, endIndex) << endl; ` ` `  `    ``startIndex = 1, endIndex = 5; ` `    ``cout << ``"Maximum Length Correct Bracket"` `            ``" Subsequence between "` `         ``<< startIndex << ``" and "` `<< endIndex << ``" = "` `         ``<< query(BOP, BCP, startIndex, endIndex) << endl; ` ` `  `    ``return` `0; ` `} `

Output:

```Maximum Length Correct Bracket Subsequence between 5 and 11 = 6
Maximum Length Correct Bracket Subsequence between 4 and 5 = 2
Maximum Length Correct Bracket Subsequence between 1 and 5 = 2
```

Time complexity for each query is O(1).

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