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 : 0Approach: 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 <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 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 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 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 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 |
Python3
# Python3 code to answer the query in constant time'''BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"'''# Function for precomputationdef constructBlanceArray(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 codeif __name__=='__main__':
string = "())(())(())(";
n = len(string)
BCP = [0 for i in range(n + 1)];
BOP = [0 for i in range(n + 1)];
constructBlanceArray(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 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 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];
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 |
Javascript
<script>// Javascript code to answer the query in constant time/*BOP[] stands for "Balanced open parentheses" BCP[] stands for "Balanced close parentheses"*/// function for precomputationfunction constructBlanceArray(BOP, BCP, str, n)
{ // Create a stack and push -1 as initial index to it.
var stk = [];
// Initialize result
var result = 0;
// Traverse all characters of given string
for (var 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.length!=0) {
BCP[i] = 1;
BOP[stk[stk.length-1]] = 1;
stk.pop();
}
// If stack is empty.
else
BCP[i] = 0;
}
}
for (var i = 1; i < n; i++) {
BCP[i] += BCP[i - 1];
BOP[i] += BOP[i - 1];
}
}// Function return output of each query in O(1)function 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 program to test above functionvar str = "())(())(())(";
var n = str.length;
var BCP = Array(n+1).fill(0);
var BOP = Array(n+1).fill(0);
constructBlanceArray(BOP, BCP, str, n);var startIndex = 5, endIndex = 11;
document.write( "Maximum Length Correct Bracket"+
" Subsequence between "
+ startIndex + " and " + endIndex + " = "
+ query(BOP, BCP, startIndex, endIndex) + "<br>");;
startIndex = 4, endIndex = 5;document.write( "Maximum Length Correct Bracket"+
" Subsequence between "
+ startIndex + " and " + endIndex + " = "
+ query(BOP, BCP, startIndex, endIndex) + "<br>");;
startIndex = 1, endIndex = 5;document.write( "Maximum Length Correct Bracket"+
" Subsequence between "
+ startIndex + " and " + endIndex + " = "
+ query(BOP, BCP, startIndex, endIndex) + "<br>");;
</script> |
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).
