给定一个由‘(‘, ‘)’, ‘[‘和‘]’组成的字符串S ，任务是通过删除有效括号的子序列来找到字符串中剩余字符的最小计数。
例子：
Input: S = “[]])([”
Output: 4
Explanation:
Removing the subsequence { str[0], str[1] } modifies S to “])([“.
Therefore, the required output is 4.
Input: S = “([)(])”
Output: 0
Explanation:
Removing the subsequence { str[0], str[2] } modifies S to “[(])”.
Removing the subsequence { str[0], str[2] } modifies S to “()”.
Removing the subsequence { str[0], str[1] } modifies S to “”.
Therefore, the required output is 0.
方法：使用Stack可以解决这个问题。请按照以下步骤解决问题：
- 这个想法是在两个单独的堆栈中处理圆括号“()”和方括号“[]” 。
- 初始化两个变量，比如 roundCount和squareCount，以分别将左括号的计数存储在‘()’和‘[]’ 的有效括号中。
- 迭代给定字符串的每个字符，并使用两个不同的堆栈计算‘()’和‘[]’的有效括号的长度。
- 最后，打印(N – 2 * (roundCount + squareCount)) 的值。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to find the minimum count of remaining
// characters left into the string by removing
// the valid subsequences
void deleteSubseq(string s)
{
// Length of the string
int N = s.size();
// Stores opening parenthesis
// '(' of the given string
stack roundStk;
// Stores square parenthesis
// '[' of the given string
stack squareStk;
// Stores count of opening parenthesis '('
// in valid subsequences
int roundCount = 0;
// Stores count of opening parenthesis '['
// in valid subsequences
int squareCount = 0;
// Iterate over each
// characters of S
for (int i = 0; i < N; i++)
{
// If current character is '['
if (s[i] == '[')
{
// insert into stack
squareStk.push(s[i]);
}
// If i is equal to ']'
else if (s[i] == ']')
{
// If stack is not empty and
// top element of stack is '['
if (squareStk.size() != 0
&& squareStk.top() == '[')
{
// Remove top element from stack
squareStk.pop();
// Update squareCount
squareCount += 1;
}
}
// If current character is '('
else if (s[i] == '(')
{
// Insert into stack
roundStk.push(s[i]);
}
// If i is equal to ')'
else
{
// If stack is not empty and
// top element of stack is '('
if (roundStk.size() != 0
&& squareStk.top() == '(')
{
// Remove top element from stack
squareStk.pop();
// Update roundCount
roundCount += 1;
}
}
}
// Print the minimum number of remaining
// characters left into S
cout << (N - (2 * squareCount + 2 * roundCount));
}
// Driver code
int main()
{
// input string
string s = "[]])([";
// function call
deleteSubseq(s);
}
// This code is contributed by gauravrajput1 Java
/*package whatever //do not write package name here */
// Java program for the above approach
import java.io.*;
import java.util.Stack;
class GFG
{
// Function to find the minimum count of remaining
// characters left into the string by removing
// the valid subsequences
public static void deleteSubseq(String s)
{
// Length of the string
int N = s.length();
// Stores opening parenthesis
// '(' of the given string
Stack roundStk = new Stack<>();
// Stores square parenthesis
// '[' of the given string
Stack squareStk = new Stack<>();
// Stores count of opening parenthesis '('
// in valid subsequences
int roundCount = 0;
// Stores count of opening parenthesis '['
// in valid subsequences
int squareCount = 0;
// Iterate over each
// characters of S
for (int i = 0; i < N; i++)
{
// If current character is '['
if (s.charAt(i) == '[')
{
// insert into stack
squareStk.push(s.charAt(i));
}
// If i is equal to ']'
else if (s.charAt(i) == ']')
{
// If stack is not empty and
// top element of stack is '['
if (squareStk.size() != 0
&& squareStk.peek() == '[')
{
// Remove top element from stack
squareStk.pop();
// Update squareCount
squareCount += 1;
}
}
// If current character is '('
else if (s.charAt(i) == '(')
{
// Insert into stack
roundStk.push(s.charAt(i));
}
// If i is equal to ')'
else
{
// If stack is not empty and
// top element of stack is '('
if (roundStk.size() != 0
&& squareStk.peek() == '(')
{
// Remove top element from stack
squareStk.pop();
// Update roundCount
roundCount += 1;
}
}
}
// Print the minimum number of remaining
// characters left into S
System.out.println(
N - (2 * squareCount + 2 * roundCount));
}
// Driver code
public static void main(String[] args)
{
// input string
String s = "[]])([";
// function call
deleteSubseq(s);
}
}
// This code is contributed by aditya7409 Python3
# Python program for the above approach
# Function to find the minimum count of remaining
# characters left into the string by removing
# the valid subsequences
def deleteSubseq(S):
# Length of the string
N = len(S)
# Stores opening parenthesis
# '(' of the given string
roundStk = []
# Stores square parenthesis
# '[' of the given string
squareStk = []
# Stores count of opening parenthesis '('
# in valid subsequences
roundCount = 0
# Stores count of opening parenthesis '['
# in valid subsequences
squareCount = 0
# Iterate over each
# characters of S
for i in S:
# If current character is '['
if i == '[':
# Insert into stack
squareStk.append(i)
# If i is equal to ']'
elif i == ']':
# If stack is not empty and
# top element of stack is '['
if squareStk and squareStk[-1] == '[':
# Remove top element from stack
squareStk.pop()
# Update squareCount
squareCount += 1
# If current character is '('
elif i == '(':
# Insert into stack
roundStk.append(i)
else:
# If stack is not empty and
# top element of stack is '('
if roundStk and roundStk[-1] == '(':
# Remove top element from stack
roundStk.pop()
# Update roundCount
roundCount += 1
# Print the minimum number of remaining
# characters left into S
print(N - (2 * squareCount + 2 * roundCount))
# Driver Code
if __name__ == '__main__':
# Given string
S = '[]])(['
# Function Call
deleteSubseq(S)C#
/*package whatever //do not write package name here */
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// Function to find the minimum count of remaining
// characters left into the string by removing
// the valid subsequences
public static void deleteSubseq(String s)
{
// Length of the string
int N = s.Length;
// Stores opening parenthesis
// '(' of the given string
Stack roundStk = new Stack();
// Stores square parenthesis
// '[' of the given string
Stack squareStk = new Stack();
// Stores count of opening parenthesis '('
// in valid subsequences
int roundCount = 0;
// Stores count of opening parenthesis '['
// in valid subsequences
int squareCount = 0;
// Iterate over each
// characters of S
for (int i = 0; i < N; i++)
{
// If current character is '['
if (s[i] == '[')
{
// insert into stack
squareStk.Push(s[i]);
}
// If i is equal to ']'
else if (s[i] == ']')
{
// If stack is not empty and
// top element of stack is '['
if (squareStk.Count != 0
&& squareStk.Peek() == '[')
{
// Remove top element from stack
squareStk.Pop();
// Update squareCount
squareCount += 1;
}
}
// If current character is '('
else if (s[i] == '(')
{
// Insert into stack
roundStk.Push(s[i]);
}
// If i is equal to ')'
else
{
// If stack is not empty and
// top element of stack is '('
if (roundStk.Count != 0
&& squareStk.Peek() == '(')
{
// Remove top element from stack
squareStk.Pop();
// Update roundCount
roundCount += 1;
}
}
}
// Print the minimum number of remaining
// characters left into S
Console.WriteLine(
N - (2 * squareCount + 2 * roundCount));
}
// Driver code
public static void Main(String[] args)
{
// input string
String s = "[]])([";
// function call
deleteSubseq(s);
}
}
// This code is contributed by 29AjayKumar 输出：
4时间复杂度： O(N)
辅助空间： O(N)
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