最长正确括号子序列集的范围查询 | 2
给定一个括号序列，或者换句话说，一个长度为 n 的字符串S，由字符'(' 和 ')' 组成。查找给定查询范围内序列的最大正确括号子序列的长度。注意：正确的括号序列是具有匹配的括号对或包含另一个嵌套的正确括号序列的序列。例如 (), (()), ()() 是一些正确的括号序列。
例子：
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方法：在上一篇文章（SET 1）中，我们讨论了每个查询在 O(long) 中工作的解决方案，现在是这篇文章，我们将看到每个查询在 O(1) 中工作的解决方案。
这个想法是基于最长有效平衡子串的 Post 长度 如果我们在一个临时数组中标记所有平衡括号/括号的索引（这里我们将其命名为 BCP[], BOP[] ）然后我们在 O(1 ） 时间。
算法 ：
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]现在您可以在 O(1) 时间内回答每个查询
(BCP[e] - BOP[s-1]])*2;下面是上述思想的实现。
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 constructBlanceArray(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 };
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 time
import java.util.*;
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 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];
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 precomputation
def 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 code
if __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 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 constructBlanceArray(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];
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
输出：
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每个查询的时间复杂度为 O(1)。