完成常规括号序列所需的右括号数

给定一个不完整的括号序列S。任务是找到使它成为规则括号序列所需的右括号’)’的数量，并打印完整的括号序列。您只能在给定括号序列的末尾添加括号。如果无法完成括号顺序，请打印“不可能”。

让我们通过以下方式定义常规的括号序列：

• 空字符串是规则的括号序列。
• 如果s是常规方括号序列，则(s)是常规方括号序列。
• 如果s＆t是正则括号序列，则st是正则括号序列。

例子：

Input : str = “(()(()(”
Output : (()(()()))
Explanation : The minimum number of ) needed to make the sequence regular are 3 which are appended at the end.

Input : str = “())(()”
Output : IMPOSSIBLE

我们需要添加最少数量的右方括号’)’，因此我们将计算不平衡的右方括号的数量，然后添加该右方括号的数量。如果在任何时候，右方括号的数目大于右方括号的数目，那么答案是不可能的。

下面是上述方法的实现：

C++

// C++ program to find number of closing
// brackets needed and complete a regular
// bracket sequence
#include 
using namespace std;
  
// Function to find number of closing
// brackets and complete a regular
// bracket sequence
void completeSuquence(string s)
{
    // Finding the length of sequence
    int n = s.length();
  
    int open = 0, close = 0;
  
    for (int i = 0; i < n; i++)
    {
        // Counting opening brackets
        if (s[i] == '(')
            open++;
        else
            // Counting closing brackets
            close++;
  
        // Checking if at any position the
        // number of closing bracket
        // is more then answer is impossible
        if (close > open)
        {
            cout << "Impossible" << endl;
            return;
        }
    }
  
    // If possible, print 's' and 
    // required closing brackets.
    cout << s;
    for (int i = 0; i < open - close; i++)
        cout << ')';
    cout << endl;
}
  
// Driver code
int main()
{
    string s = "(()(()(";
    completeSuquence(s);
    return 0;
}
  
// This code is contributed by
// sanjeev2552

Java

// Java program to find number of closing
// brackets needed and complete a regular
// bracket sequence
class GFG {
  
    // Function to find number of closing
    // brackets and complete a regular
    // bracket sequence
    static void completeSequence(String s)
    {
        // Finding the length of sequence
        int n = s.length();
  
        int open = 0, close = 0;
  
        for (int i = 0; i < n; i++) {
  
            // Counting opening brackets
            if (s.charAt(i) == '(')
                open++;
            else
                // Counting closing brackets
                close++;
  
            // Checking if at any position the
            // number of closing bracket
            // is more then answer is impossible
            if (close > open) {
                System.out.print("IMPOSSIBLE");
                return;
            }
        }
  
        // If possible, print 's' and required closing
        // brackets.
        System.out.print(s);
        for (int i = 0; i < open - close; i++)
            System.out.print(")");          
    }
  
    // Driver code
    public static void main(String[] args)
    {
        String s = "(()(()(";
        completeSequence(s);
    }
}

Python 3

# Python 3 program to find number of 
# closing brackets needed and complete 
# a regular bracket sequence
  
# Function to find number of closing
# brackets and complete a regular
# bracket sequence
def completeSequence(s):
  
    # Finding the length of sequence
    n = len(s)
  
    open = 0
    close = 0
  
    for i in range(n):
  
        # Counting opening brackets
        if (s[i] == '('):
            open += 1
        else:
              
            # Counting closing brackets
            close += 1
  
        # Checking if at any position the
        # number of closing bracket
        # is more then answer is impossible
        if (close > open):
            print("IMPOSSIBLE")
            return
  
    # If possible, print 's' and 
    # required closing brackets.
    print(s, end = "")
    for i in range(open - close):
        print(")", end = "")
  
# Driver code
if __name__ == "__main__":
      
    s = "(()(()("
    completeSequence(s)
  
# This code is contributed by ita_c

C#

// C# program to find number of closing 
// brackets needed and complete a 
// regular bracket sequence 
using System;
  
class GFG
{
// Function to find number of closing
// brackets and complete a regular
// bracket sequence
static void completeSequence(String s)
{
    // Finding the length of sequence
    int n = s.Length;
  
    int open = 0, close = 0;
  
    for (int i = 0; i < n; i++) 
    {
  
        // Counting opening brackets
        if (s[i] == '(')
            open++;
        else
            // Counting closing brackets
            close++;
  
        // Checking if at any position the
        // number of closing bracket
        // is more then answer is impossible
        if (close > open) 
        {
            Console.Write("IMPOSSIBLE");
            return;
        }
    }
  
    // If possible, print 's' and 
    // required closing brackets.
    Console.Write(s);
    for (int i = 0; i < open - close; i++)
        Console.Write(")");         
}
  
// Driver Code
static void Main()
{
    String s = "(()(()(";
    completeSequence(s);
}
}
  
// This code is contributed
// by ANKITRAI1

PHP

 $open) 
        { 
            echo ("IMPOSSIBLE"); 
            return; 
        } 
    } 
  
    // If possible, print 's' and 
    // required closing brackets. 
    echo ($s); 
    for ($i = 0; $i < $open - $close; $i++) 
        echo (")");     
} 
  
// Driver Code 
$s = "(()(()("; 
completeSequence($s); 
  
// This code is contributed 
// by ajit 
?>

输出：

(()(()()))