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。任务是找到使其成为常规括号序列所需的右括号 ‘)’ 的数量并打印完整的括号序列。您只能在给定的括号序列的末尾添加括号。如果无法完成括号序列，请打印“IMPOSSIBLE”。
让我们以下列方式定义一个正则括号序列：

• 空字符串是一个常规的括号序列。
• 如果 s 是正则括号序列，则 (s) 是正括号序列。
• 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括号 ‘)’，因此我们将计算不平衡的左括号的数量，然后我们将添加该数量的右括号。如果在任何时候结束括号的数量大于开始括号，那么答案是IMPOSSIBLE 。
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
?>

Javascript

    // Javascript 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
    function completeSequence(s)
    {
        // Finding the length of sequence
        let n = s.length;
 
        let open = 0, close = 0;
 
        for (let 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)
            {
                document.write("IMPOSSIBLE");
                return;
            }
        }
 
        // If possible, print 's' and
        // required closing brackets.
        document.write(s);
        for (let i = 0; i < open - close; i++)
            document.write(")");        
    }
     
    let s = "(()(()(";
    completeSequence(s);
         

输出：

(()(()()))

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