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 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.
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 the above idea.
C++
// CPP code to answer the query in constant time#include <bits/stdc++.h>using namespace std;/*BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"*/// function for precomputationvoid 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 functionint 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;} |
Java
// Java code to answer the query in constant timeimport java.util.*;class GFG{/*BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"*/// Function for precomputationstatic void constructBlanceArray(int BOP[], int BCP[], String str, int n){ // Create a stack and push -1 // as initial index to it. Stack<Integer> 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 indexs 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 codepublic static void main(String[] args){ String str = "())(())(())("; int n = str.length(); int BCP[] = new int[n + 1]; int BOP[] = new int[n + 1]; constructBlanceArray(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 |
C#
// C# code to answer the query// in constant timeusing System;using System.Collections.Generic;class GFG{ /* BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses" */ // Function for precomputation static void constructBlanceArray(int[] BOP, int[] BCP, String str, int n) { // Create a stack and push -1 // as initial index to it. Stack<int> stk = new Stack<int>();; // 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 indexs 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]; constructBlanceArray(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 |
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).
Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.
