Number of closing brackets needed to complete a regular bracket sequence

28 Jun, 2022
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  
 
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.Below is the implementation of the above approach:  
C++
Java
Python 3
C#
PHP
Javascript
// 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 sequencevoid 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 codeint 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 sequenceclass 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 sequencedef 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 codeif __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 sequenceusing System; class GFG{// Function to find number of closing// brackets and complete a regular// bracket sequencestatic 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 Codestatic 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 sequencefunction 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>
(()(()()))
 
ankthon
jit_t
ukasp
sanjeev2552
divyesh072019
rkbhola5
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