Number of closing brackets needed to complete a regular bracket sequence

Given an incomplete bracket sequence S. The task is to find the number of closing brackets ')' needed to make it a regular bracket sequence and print the complete bracket sequence. You are allowed to add the brackets only at the end of the given bracket sequence. If it is not possible to complete the bracket sequence, print "IMPOSSIBLE".

Let us define a regular bracket sequence in the following way: 

• Empty string is a regular bracket sequence.
• If s is a regular bracket sequence, then (s) is a regular bracket sequence.
• If s & t are regular bracket sequences, then st is a regular bracket sequence.

Examples:  

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 

Approach :

We need to add minimal number of closing brackets ')', so we will count the number of unbalanced opening brackets and then we will add that amount of closing brackets. If at any point the number of the closing bracket is greater than the opening bracket then the answer is IMPOSSIBLE.

Algorithm :

1. Create two variables open = 0 and close = 0
2. Traverse on a string from i = 0 to i = n(size of string)
3. If current element is '(' then increment open else if current element is ')' then increment close.
4. While traversing check if count of close is greater than open or not if yes then print Impossible return to main 
5. After completion of traversal calculate (open-close) as that many times closing brackets required to make the sequence balanced.

Below is the implementation of the above approach: 

// C++ program to find number of closing
// brackets needed and complete a regular
// bracket sequence
#include <iostream>
using namespace std;

// Function to find number of closing
// brackets and complete a regular
// bracket sequence
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)
        {
            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 = "(()(()(";
    completeSequence(s);
    return 0;
}

// This code is contributed by
// sanjeev2552

// 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 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# 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
// PHP 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 
    $n = strlen($s); 
    $open = 0;
    $close = 0; 

    for ($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) 
        { 
            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 
?>

<script>
    // 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);
        
</script>

(()(()()))

Complexity Analysis:

• Time Complexity: O(n) , where n is size of given string
• Auxiliary Space: O(1) , as we are not using any extra space.