# 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 â€˜(â€˜ and â€˜)â€™. 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 go to see a solution that works in O(1) for each query.

The idea is based on the Post length of the longest valid balanced substring If we marked indexes of all Balanced parentheses/brackets 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.
Traverse 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 the above idea.

## C++

 // 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 constructBalanceArray(int BOP[], int BCP[],                           char* str, int n) {       // Create a stack and push -1 as initial index to it.     stack 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 indexes 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 };       constructBalanceArray(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; }

## Java

 // Java code to answer the query in constant time import java.util.*;   class GFG{   /* BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"   */   // Function for precomputation static void constructBalanceArray(int BOP[], int BCP[],                                 String str, int n) {           // Create a stack and push -1     // as initial index to it.     Stack stk = new Stack<>();;       // Traverse all characters of given String     for(int i = 0; i < n; i++)     {                   // If opening bracket, push index of it         if (str.charAt(i) == '(')             stk.add(i);                       // If closing bracket, i.e., str[i] = ')'         else         {                           // If closing bracket, i.e., str[i] = ')'             // && stack is not empty then mark both             // "open & close" bracket indexes as 1 .             // Pop the previous opening bracket's index             if (!stk.isEmpty())             {                 BCP[i] = 1;                 BOP[stk.peek()] = 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) static 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 code public static void main(String[] args) {       String str = "())(())(())(";     int n = str.length();       int BCP[] = new int[n + 1];     int BOP[] = new int[n + 1];       constructBalanceArray(BOP, BCP, str, n);       int startIndex = 5, endIndex = 11;     System.out.print("Maximum Length Correct " +                      "Bracket Subsequence between " +                      startIndex + " and " + endIndex +                      " = " + query(BOP, BCP, startIndex,                                    endIndex) + "\n");       startIndex = 4;     endIndex = 5;     System.out.print("Maximum Length Correct " +                       "Bracket Subsequence between " +                      startIndex + " and " + endIndex +                      " = " + query(BOP, BCP, startIndex,                                    endIndex) + "\n");       startIndex = 1;     endIndex = 5;     System.out.print("Maximum Length Correct " +                      "Bracket Subsequence between " +                      startIndex + " and " + endIndex +                      " = " + query(BOP, BCP, startIndex,                                    endIndex) + "\n"); } }   // This code is contributed by 29AjayKumar

## Python3

 # Python3 code to answer the query in constant time   ''' BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"   ''' # Function for precomputation def constructBalanceArray(BOP, BCP, str, n):           # Create a stack and push -1     # as initial index to it.     stk = []       # Traverse all characters of given String     for i in range(n):                   # If opening bracket, push index of it         if (str[i] == '('):             stk.append(i);                       # If closing bracket, i.e., str[i] = ')'         else:                           # If closing bracket, i.e., str[i] = ')'             # && stack is not empty then mark both             # "open & close" bracket indexes as 1 .             # Pop the previous opening bracket's index             if (len(stk) != 0):                 BCP[i] = 1;                 BOP[stk[-1]] = 1;                 stk.pop();               # If stack is empty.             else:                 BCP[i] = 0;               for i in range(1, n):           BCP[i] += BCP[i - 1];         BOP[i] += BOP[i - 1];       # Function return output of each query in O(1) def query(BOP, BCP, s, e):       if (BOP[s - 1] == BOP[s]):         return (BCP[e] - BOP[s]) * 2;           else:         return (BCP[e] - BOP[s] + 1) * 2;   # Driver code if __name__=='__main__':       string = "())(())(())(";     n = len(string)       BCP = [0 for i in range(n + 1)];     BOP = [0 for i in range(n + 1)];       constructBalanceArray(BOP, BCP, string, n);     startIndex = 5     endIndex = 11;     print("Maximum Length Correct " +                      "Bracket Subsequence between " +                      str(startIndex) + " and " + str(endIndex) +                      " = " + str(query(BOP, BCP, startIndex,                                    endIndex)));     startIndex = 4;     endIndex = 5;     print("Maximum Length Correct " +                       "Bracket Subsequence between " +                      str(startIndex) + " and " + str(endIndex) +                      " = " + str(query(BOP, BCP, startIndex,                                    endIndex)))     startIndex = 1;     endIndex = 5;     print("Maximum Length Correct " +                      "Bracket Subsequence between " +                      str(startIndex) + " and " + str(endIndex) +                      " = " + str(query(BOP, BCP, startIndex,                                    endIndex)));   # This code is contributed by rutvik_56.

## C#

 // C# code to answer the query // in constant time using System; using System.Collections.Generic; class GFG{       /*     BOP[] stands for "Balanced open parentheses"     BCP[] stands for "Balanced close parentheses"     */       // Function for precomputation     static void constructBalanceArray(int[] BOP, int[] BCP,                                      String str, int n)     {           // Create a stack and push -1         // as initial index to it.         Stack stk = new Stack();;           // 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);               // If closing bracket, i.e., str[i] = ')'             else             {                   // If closing bracket, i.e., str[i] = ')'                 // && stack is not empty then mark both                 // "open & close" bracket indexes as 1 .                 // Pop the previous opening bracket's index                 if (stk.Count != 0)                 {                     BCP[i] = 1;                     BOP[stk.Peek()] = 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)     static 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 code     public static void Main(String[] args)     {         String str = "())(())(())(";         int n = str.Length;         int[] BCP = new int[n + 1];         int[] BOP = new int[n + 1];         constructBalanceArray(BOP, BCP, str, n);         int startIndex = 5, endIndex = 11;         Console.Write("Maximum Length Correct " +                       "Bracket Subsequence between " +                        startIndex + " and " + endIndex + " = " +                        query(BOP, BCP, startIndex, endIndex) + "\n");           startIndex = 4;         endIndex = 5;         Console.Write("Maximum Length Correct " +                       "Bracket Subsequence between " +                        startIndex + " and " + endIndex + " = " +                        query(BOP, BCP, startIndex, endIndex) + "\n");           startIndex = 1;         endIndex = 5;         Console.Write("Maximum Length Correct " +                       "Bracket Subsequence between " +                        startIndex + " and " + endIndex + " = " +                        query(BOP, BCP, startIndex, endIndex) + "\n");     } }   // This code is contributed by Amit Katiyar

## Javascript



Output

Maximum Length Correct Bracket Subsequence between 5 and 11 = 4
Maximum Length Correct Bracket Subsequence between 4 and 5 = 0
Maximum Length Correct Bracket Subsequence between 1 and 5 = 2

The time complexity for each query is O(1).
Overall time Complexity: O(n)
Auxiliary Space: O(n)

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