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 = ())(())(())(
Output: 10
Explanation: Longest Correct Bracket Subsequence is ()(())(())Input: S = ())(())(())(0
Output: 0
Range Queries for Longest Correct Bracket Subsequence using segment trees:
To solve the problem follow the below idea:
Segment Trees can be used to solve this problem efficiently At each node of the segment tree, we store the following:
- Number of correctly matched pairs of brackets
- Number of unused open brackets
- Number of unused closed brackets
(unused open bracket – means they can’t be matched with any closing bracket, for e.g S = )( contains an unused open and an unused closed bracket) For each interval [L, R], we can match X number of unused open brackets ‘(‘ in interval [L, MID] with unused closed brackets ‘)’ in interval [MID + 1, R] where X = minimum(number of unused ‘(‘ in [L, MID], number of unused ‘)’ in [MID + 1, R]) Hence, X is also the number of correctly matched pairs built by combination. So, for interval [L, R]
- Total number of correctly matched pairs becomes the sum of correctly matched pairs in left child and correctly matched pairs in right child and number of combinations of unused ‘(‘ and unused ‘)’ from left and right child respectively
a[L, R] = a[L, MID] + a[MID + 1, R] + X
- Total number of unused open brackets becomes the sum of unused open brackets in left child and unused open brackets in right child minus X (minus – because we used X unused ‘(‘ from left child to match with unused ‘) from right child)
a[L, R] = b[L, MID] + b[MID + 1, R] – X
- Similarly, for unused closed brackets, following relation holds
a[L, R] = c[L, MID] + c[MID + 1, R] – X
where a, b and c are the representations described above for each node to be stored in
Below is the implementation of the above approach:
C++
/* CPP Program to find the longest correctbracket subsequence in a given range */#include <bits/stdc++.h>using namespace std;/* Declaring Structure for storingthree values in each segment tree node */struct Node { int pairs; int open; // unused int closed; // unused Node() { pairs = open = closed = 0; }};// A utility function to get the middle index from corner// indexes.int getMid(int s, int e) { return s + (e - s) / 2; }// Returns Parent Node after merging its left and right// childNode merge(Node leftChild, Node rightChild){ Node parentNode; int minMatched = min(leftChild.open, rightChild.closed); parentNode.pairs = leftChild.pairs + rightChild.pairs + minMatched; parentNode.open = leftChild.open + rightChild.open - minMatched; parentNode.closed = leftChild.closed + rightChild.closed - minMatched; return parentNode;}// A recursive function that constructs Segment Tree// for string[ss..se]. si is index of current node in// segment tree stvoid constructSTUtil(char str[], int ss, int se, Node* st, int si){ // If there is one element in string, store it in // current node of segment tree and return if (ss == se) { // since it contains one element, pairs // will be zero st[si].pairs = 0; // check whether that one element is opening // bracket or not st[si].open = (str[ss] == '(' ? 1 : 0); // check whether that one element is closing // bracket or not st[si].closed = (str[ss] == ')' ? 1 : 0); return; } // If there are more than one elements, then recur // for left and right subtrees and store the relation // of values in this node int mid = getMid(ss, se); constructSTUtil(str, ss, mid, st, si * 2 + 1); constructSTUtil(str, mid + 1, se, st, si * 2 + 2); // Merge left and right child into the Parent Node st[si] = merge(st[si * 2 + 1], st[si * 2 + 2]);}/* Function to construct segment tree from givenstring. This function allocates memory for segmenttree and calls constructSTUtil() to fill theallocated memory */Node* constructST(char str[], int n){ // Allocate memory for segment tree // Height of segment tree int x = (int)(ceil(log2(n))); // Maximum size of segment tree int max_size = 2 * (int)pow(2, x) - 1; // Declaring array of structure Allocate memory Node* st = new Node[max_size]; // Fill the allocated memory st constructSTUtil(str, 0, n - 1, st, 0); // Return the constructed segment tree return st;}/* A Recursive function to get the desiredMaximum Sum Sub-Array,The following are parameters of the function-st --> Pointer to segment treesi --> Index of the segment tree Nodess & se --> Starting and ending indexes of the segment represented by current Node, i.e., tree[index]qs & qe --> Starting and ending indexes of query range */Node queryUtil(Node* st, int ss, int se, int qs, int qe, int si){ // No overlap if (ss > qe || se < qs) { // returns a Node for out of bounds condition Node nullNode; return nullNode; } // Complete overlap if (ss >= qs && se <= qe) { return st[si]; } // Partial Overlap Merge results of Left // and Right subtrees int mid = getMid(ss, se); Node left = queryUtil(st, ss, mid, qs, qe, si * 2 + 1); Node right = queryUtil(st, mid + 1, se, qs, qe, si * 2 + 2); // merge left and right subtree query results Node res = merge(left, right); return res;}/* Returns the maximum length correct bracketsubsequencebetween start and endIt mainly uses queryUtil(). */int query(Node* st, int qs, int qe, int n){ Node res = queryUtil(st, 0, n - 1, qs, qe, 0); // since we are storing numbers pairs // and have to return maximum length, hence // multiply no of pairs by 2 return 2 * res.pairs;}// Driver Codeint main(){ char str[] = "())(())(())("; int n = strlen(str); // Build segment tree from given string Node* st = constructST(str, n); // Function call int startIndex = 0, endIndex = 11; cout << "Maximum Length Correct Bracket" " Subsequence between " << startIndex << " and " << endIndex << " = " << query(st, startIndex, endIndex, n) << endl; startIndex = 1, endIndex = 2; cout << "Maximum Length Correct Bracket" " Subsequence between " << startIndex << " and " << endIndex << " = " << query(st, startIndex, endIndex, n) << endl; return 0;} |
Java
// Java Program to find the longest correct bracket // subsequence in a given range import java.util.*;import java.lang.*;import java.io.*;// Declaring Structure for storing three values in each segment tree nodeclass Node { int pairs; int open; // unused int closed; // unused Node() { pairs = 0; open = 0; closed = 0; }}class SegmentTree { Node[] st; int n; // A utility function to get the middle index from corner indexes. int getMid(int s, int e) { return s + (e - s) / 2; } // Returns Parent Node after merging its left and right child Node merge(Node leftChild, Node rightChild) { Node parentNode = new Node(); int minMatched = Math.min(leftChild.open, rightChild.closed); parentNode.pairs = leftChild.pairs + rightChild.pairs + minMatched; parentNode.open = leftChild.open + rightChild.open - minMatched; parentNode.closed = leftChild.closed + rightChild.closed - minMatched; return parentNode; } // A recursive function that constructs Segment Tree for string[ss..se]. // si is index of current node in segment tree st void constructSTUtil(String str, int ss, int se, int si) { // If there is one element in string, store it in current node of // segment tree and return if (ss == se) { // since it contains one element, pairs will be zero st[si].pairs = 0; // check whether that one element is opening bracket or not st[si].open = (str.charAt(ss) == '(') ? 1 : 0; // check whether that one element is closing bracket or not st[si].closed = (str.charAt(ss) == ')') ? 1 : 0; return; } // If there are more than one elements, then recur for left and right // subtrees and store the relation of values in this node int mid = getMid(ss, se); constructSTUtil(str, ss, mid, si * 2 + 1); constructSTUtil(str, mid + 1, se, si * 2 + 2); // Merge left and right child into the Parent Node st[si] = merge(st[si * 2 + 1], st[si * 2 + 2]); } // Function to construct segment tree from given string. // This function allocates memory for segment tree and calls // constructSTUtil() to fill the allocated memory void constructST(String str) { // Allocate memory for segment tree // Height of segment tree int x = (int) (Math.ceil(Math.log(n) / Math.log(2))); // Maximum size of segment tree int max_size = 2 * (int) Math.pow(2, x) - 1; st = new Node[max_size]; for (int i = 0; i < max_size; i++) st[i] = new Node(); // Fill the allocated memory st constructSTUtil(str, 0, n - 1, 0); } // A Recursive function to get the desired Maximum Sum Sub-Array, // The following are parameters of the function- // st --> Pointer to segment tree // si --> Index of the segment tree Node // ss & se --> Starting and ending indexes of the segment // represented by current Node, i.e., tree[index] // qs & qe --> Starting and ending indexes of query range Node queryUtil(int ss, int se, int qs, int qe, int si) { // No overlap if (ss > qe || se < qs) // returns a Node for out of bounds condition return new Node(); // Complete overlap if (ss >= qs && se <= qe) return st[si]; // Partial Overlap Merge results of Left and Right subtrees int mid = getMid(ss, se); Node left = queryUtil(ss, mid, qs, qe, si * 2 + 1); Node right = queryUtil(mid + 1, se, qs, qe, si * 2 + 2); return merge(left, right); } // The function to get the maximum length of correct bracket subsequence // for given range. The following are parameters of the function- // st --> Pointer to segment tree // qs & qe --> Starting and ending indexes of query range int query(int qs, int qe) { Node node = queryUtil(0, n - 1, qs, qe, 0); return 2 * node.pairs; } // Driver code public static void main(String args[]) { String str = "())(())(())("; int n = str.length(); SegmentTree tree = new SegmentTree(); tree.n = n; tree.constructST(str); int qs = 0; int qe = n - 1; System.out.println("Maximum Length Correct Bracket Subsequence between " + qs + " and " + qe + " = " + tree.query(qs, qe)); qs = 1; qe = 2; System.out.println("Maximum Length Correct Bracket Subsequence between " + qs + " and " + qe + " = " + tree.query(qs, qe)); }} |
Python3
# Python Program to find the longest correct bracket # subsequence in a given range import math# Declaring Structure for storing three values in each segment tree nodeclass Node: def __init__(self): self.pairs = 0 self.open = 0 # unused self.closed = 0 # unused# A utility function to get the middle index from corner indexes.def getMid(s: int, e: int) -> int: return s + (e - s) // 2# Returns Parent Node after merging its left and right childdef merge(leftChild: Node, rightChild: Node) -> Node: parentNode = Node() minMatched = min(leftChild.open, rightChild.closed) parentNode.pairs = leftChild.pairs + rightChild.pairs + minMatched parentNode.open = leftChild.open + rightChild.open - minMatched parentNode.closed = leftChild.closed + rightChild.closed - minMatched return parentNode# A recursive function that constructs Segment Tree for string[ss..se].# si is index of current node in segment tree stdef constructSTUtil(str: str, ss: int, se: int, st: list, si: int): # If there is one element in string, store it in current node of segment tree and return if ss == se: # since it contains one element, pairs will be zero st[si].pairs = 0 # check whether that one element is opening bracket or not st[si].open = 1 if str[ss] == '(' else 0 # check whether that one element is closing bracket or not st[si].closed = 1 if str[ss] == ')' else 0 return # If there are more than one elements, then recur for left and right subtrees # and store the relation of values in this node mid = getMid(ss, se) constructSTUtil(str, ss, mid, st, si * 2 + 1) constructSTUtil(str, mid + 1, se, st, si * 2 + 2) # Merge left and right child into the Parent Node st[si] = merge(st[si * 2 + 1], st[si * 2 + 2])# Function to construct segment tree from given string.# This function allocates memory for segment tree and calls constructSTUtil()# to fill the allocated memorydef constructST(str: str, n: int) -> list: # Allocate memory for segment tree # Height of segment tree x = int(math.ceil(math.log2(n))) # Maximum size of segment tree max_size = 2 * int(2 ** x) - 1 # Declaring array of structure Allocate memory st = [Node() for _ in range(max_size)] # Fill the allocated memory st constructSTUtil(str, 0, n - 1, st, 0) # Return the constructed segment tree return st# A Recursive function to get the desired Maximum Sum Sub-Array,# The following are parameters of the function-# st --> Pointer to segment tree# si --> Index of the segment tree Node# ss & se --> Starting and ending indexes of the segment # represented by current Node, i.e., tree[index]# qs & qe --> Starting and ending indexes of query rangedef queryUtil(st: list, ss: int, se: int, qs: int, qe: int, si: int) -> Node: # No overlap if ss > qe or se < qs: # returns a Node for out of bounds condition return Node() # Complete overlap if ss >= qs and se <= qe: return st[si] # Partial Overlap Merge results of Left and Right subtrees mid = getMid(ss, se) left = queryUtil(st, ss, mid, qs, qe, si * 2 + 1) right = queryUtil(st, mid + 1, se, qs, qe, si * 2 + 2) return merge(left, right)# The function to get the maximum length of correct bracket subsequence# for given range. The following are parameters of the function-# st --> Pointer to segment tree# qs & qe --> Starting and ending indexes of query rangedef query(st: list, n: int, qs: int, qe: int) -> int: node = queryUtil(st, 0, n - 1, qs, qe, 0) return 2 * node.pairs# Driver codedef main(): str = "())(())(())(" n = len(str) st = constructST(str, n) qs = 0 qe = n - 1 print("Maximum Length Correct Bracket Subsequence between", qs, "and", qe, "=", query(st, n, qs, qe)) qs = 1 qe = 2 print("Maximum Length Correct Bracket Subsequence between", qs, "and", qe, "=", query(st, n, qs, qe))if __name__ == '__main__': main()# This code is contributed by Vikram_Shirsat |
Javascript
// Javascript Program to find the longest correct bracket // subsequence in a given range class Node { constructor() { this.pairs = 0; this.open = 0; // unused this.closed = 0; // unused }}// A utility function to get the middle index from corner indexes.function getMid(s, e) { return s + Math.floor((e - s) / 2);}// Returns Parent Node after merging its left and right childfunction merge(leftChild, rightChild) { const parentNode = new Node(); const minMatched = Math.min(leftChild.open, rightChild.closed); parentNode.pairs = leftChild.pairs + rightChild.pairs + minMatched; parentNode.open = leftChild.open + rightChild.open - minMatched; parentNode.closed = leftChild.closed + rightChild.closed - minMatched; return parentNode;}// A recursive function that constructs Segment Tree for string[ss..se].// si is index of current node in segment tree stfunction constructSTUtil(str, ss, se, st, si) { // If there is one element in string, store it in current node of segment tree and return if (ss === se) { // since it contains one element, pairs will be zero st[si].pairs = 0; // check whether that one element is opening bracket or not st[si].open = str[ss] === '(' ? 1 : 0; // check whether that one element is closing bracket or not st[si].closed = str[ss] === ')' ? 1 : 0; return; } // If there are more than one elements, then recur for left and right subtrees // and store the relation of values in this node const mid = getMid(ss, se); constructSTUtil(str, ss, mid, st, si * 2 + 1); constructSTUtil(str, mid + 1, se, st, si * 2 + 2); // Merge left and right child into the Parent Node st[si] = merge(st[si * 2 + 1], st[si * 2 + 2]);}// Function to construct segment tree from given string.// This function allocates memory for segment tree and calls constructSTUtil()// to fill the allocated memoryfunction constructST(str, n) { // Allocate memory for segment tree // Height of segment tree const x = Math.ceil(Math.log2(n)); // Maximum size of segment tree const max_size = 2 * Math.pow(2, x) - 1; // Declaring array of structure Allocate memory const st = Array(max_size).fill().map(_ => new Node()); // Fill the allocated memory st constructSTUtil(str, 0, n - 1, st, 0); // Return the constructed segment tree return st;}// A Recursive function to get the desired Maximum Sum Sub-Array,// The following are parameters of the function-// st --> Pointer to segment tree// si --> Index of the segment tree Node// ss & se --> Starting and ending indexes of the segment // represented by current Node, i.e., tree[index]// qs & qe --> Starting and ending indexes of query rangefunction queryUtil(st, ss, se, qs, qe, si) { if (ss > qe || se < qs) { return new Node(); } if (ss >= qs && se <= qe) { return st[si]; } const mid = getMid(ss, se); const left = queryUtil(st, ss, mid, qs, qe, si * 2 + 1); const right = queryUtil(st, mid + 1, se, qs, qe, si * 2 + 2); return merge(left, right);}// The function to get the maximum length of correct bracket subsequence// for given range. The following are parameters of the function-// st --> Pointer to segment tree// qs & qe --> Starting and ending indexes of query rangefunction query(st, n, qs, qe) { const node = queryUtil(st, 0, n - 1, qs, qe, 0); return 2 * node.pairs;}// Driver codefunction main() { const str = "())(())(())("; const n = str.length; const st = constructST(str, n); let qs = 0; let qe = n - 1; console.log(`Maximum Length Correct Bracket Subsequence between ${qs} and ${qe} = ${query(st, n, qs, qe)}`); qs = 1; qe = 2; console.log(`Maximum Length Correct Bracket Subsequence between ${qs} and ${qe} = ${query(st, n, qs, qe)}`);}main();// This code is contributed by sdeadityasharma |
C#
// C# Program to find the longest correct bracket // subsequence in a given range using System;// Declaring Structure for storing three values in each segment tree node // A utility function to get the middle index from corner indexes.class Node { public int pairs; public int open; public int closed; public Node() { pairs = 0; open = 0; closed = 0; }}class SegmentTree { public Node[] st; public int n; // A utility function to get the middle index from corner indexes. int getMid(int s, int e) { return s + (e - s) / 2; } // Returns Parent Node after merging its left and right child Node merge(Node leftChild, Node rightChild) { Node parentNode = new Node(); int minMatched = Math.Min(leftChild.open, rightChild.closed); parentNode.pairs = leftChild.pairs + rightChild.pairs + minMatched; parentNode.open = leftChild.open + rightChild.open - minMatched; parentNode.closed = leftChild.closed + rightChild.closed - minMatched; return parentNode; } // A recursive function that constructs Segment Tree for string[ss..se]. // si is index of current node in segment tree st void constructSTUtil(string str, int ss, int se, int si) { // If there is one element in string, store it in current node of // segment tree and return if (ss == se) { // since it contains one element, pairs will be zero st[si].pairs = 0; // check whether that one element is opening bracket or not st[si].open = (str[ss] == '(') ? 1 : 0; // check whether that one element is closing bracket or not // If there are more than one elements, then recur for left and right // subtrees and store the relation of values in this node st[si].closed = (str[ss] == ')') ? 1 : 0; return; } // If there are more than one elements, then recur for left and right // subtrees and store the relation of values in this node int mid = getMid(ss, se); constructSTUtil(str, ss, mid, si * 2 + 1); constructSTUtil(str, mid + 1, se, si * 2 + 2); // Merge left and right child into the Parent Node st[si] = merge(st[si * 2 + 1], st[si * 2 + 2]); } // Function to construct segment tree from given string. // This function allocates memory for segment tree and calls // constructSTUtil() to fill the allocated memory public void constructST(string str) { // Allocate memory for segment tree // Height of segment tree int x = (int) Math.Ceiling(Math.Log(n) / Math.Log(2)); // Maximum size of segment tree int max_size = 2 * (int) Math.Pow(2, x) - 1; st = new Node[max_size]; for (int i = 0; i < max_size; i++) st[i] = new Node(); // Fill the allocated memory st constructSTUtil(str, 0, n - 1, 0); } // A Recursive function to get the desired Maximum Sum Sub-Array, // The following are parameters of the function- // st --> Pointer to segment tree // si --> Index of the segment tree Node // ss & se --> Starting and ending indexes of the segment // represented by current Node, i.e., tree[index] // qs & qe --> Starting and ending indexes of query range Node queryUtil(int ss, int se, int qs, int qe, int si) { // No overlap if (ss > qe || se < qs) // returns a Node for out of bounds condition return new Node(); if (ss >= qs && se <= qe) return st[si]; int mid = getMid(ss, se); Node left = queryUtil(ss, mid, qs, qe, si * 2 + 1); Node right = queryUtil(mid + 1, se, qs, qe, si * 2 + 2); return merge(left, right); } // The function to get the maximum length of correct bracket subsequence // for given range. The following are parameters of the function- // st --> Pointer to segment tree // qs & qe --> Starting and ending indexes of query range public int query(int qs, int qe) { Node node = queryUtil(0, n - 1, qs, qe, 0); return 2 * node.pairs; }//Driver code public static void Main(string[] args) { string str = "())(())(())("; int n = str.Length; SegmentTree tree = new SegmentTree(); tree.n = n; tree.constructST(str); int qs = 0; int qe = n - 1; Console.WriteLine("Maximum Length Correct Bracket Subsequence between " + qs + " and " + qe + " = " + tree.query(qs, qe)); qs = 1; qe = 2; Console.WriteLine("Maximum Length Correct Bracket Subsequence between " + qs + " and " + qe + " = " + tree.query(qs, qe)); }} |
Output
Maximum Length Correct Bracket Subsequence between 0 and 11 = 10 Maximum Length Correct Bracket Subsequence between 1 and 2 = 0
Time complexity: O(N*log N), where N is the size of the string
Auxiliary Space: O(N)
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