Open In App

Balance a string after removing extra brackets

Last Updated : 01 Sep, 2022
Comments
Improve
Suggest changes
Like Article
Like
Report

Given a string of characters with opening and closing brackets. The task is to remove extra brackets from string and balance it.

Examples: 

Input: str = "gau)ra)v(ku(mar(rajput))" 
Output: gaurav(ku(mar(rajput)))

Input: str = "1+5)+5+)6+(5+9)*9" 
Output: 1+5+5+6+(5+9)*9

Approach: 

  • Start traversing from left to right.
  • Check if the element at current index is an opening bracket '(' then print that bracket and increment count.
  • Check if the element at current index is a closing bracket ')' and if the count is not equal to zero then print it and decrement the count.
  • Check if there is any element other than brackets at the current index in the string then print it.
  • And in last if the count is not equal to zero then print ')' equal to the number of the count to balance the string.

Below is the implementation of above approach: 

C++ 
// C++ implementation of above approach
#include <bits/stdc++.h>
using namespace std;

// Print balanced and remove 
// extra brackets from string
void balancedString(string str)
{
    int count = 0, i;
    int n = str.length();

    // Maintain a count for opening brackets
    // Traversing string
    for (i = 0; i < n; i++) {

        // check if opening bracket
        if (str[i] == '(') {

            // print str[i] and increment count by 1
            cout << str[i];
            count++;
        }

        // check if closing bracket and count != 0
        else if (str[i] == ')' && count != 0) {
            cout << str[i];

            // decrement count by 1
            count--;
        }

        // if str[i] not a closing brackets
        // print it
        else if (str[i] != ')')
            cout << str[i];
    }

    // balanced brackets if opening brackets
    // are more then closing brackets
    if (count != 0)
        // print remaining closing brackets
        for (i = 0; i < count; i++)
           cout << ")";
}

// Driver code
int main()
{

    string str = "gau)ra)v(ku(mar(rajput))";
     balancedString(str);

    return 0;
}
Java 
// Java implementation of above approach
class GFG { 

// Print balanced and remove 
// extra brackets from string 
public static void balancedString(String str) 
{ 
    int count = 0, i; 
    int n = str.length(); 

    // Maintain a count for opening brackets 
    // Traversing string 
    for (i = 0; i < n; i++) { 

        // check if opening bracket 
        if (str.charAt(i) == '(') { 

            // print str.charAt(i) and increment count by 1 
            System.out.print(str.charAt(i)); 
            count++; 
        } 

        // check if closing bracket and count != 0 
        else if (str.charAt(i) == ')' && count != 0) { 
            System.out.print(str.charAt(i));

            // decrement count by 1 
            count--; 
        } 

        // if str.charAt(i) not a closing brackets 
        // print it 
        else if (str.charAt(i) != ')') 
            System.out.print(str.charAt(i)); 
    } 

    // balanced brackets if opening brackets 
    // are more then closing brackets 
    if (count != 0) 
        // print remaining closing brackets 
        for (i = 0; i < count; i++) 
            System.out.print(")"); 
} 

// Driver Method 
public static void main(String args[]) 
{ 
    String str = "gau)ra)v(ku(mar(rajput))"; 
    balancedString(str); 
} 
}
Python3 
# Python implementation of above approach

# Print balanced and remove 
# extra brackets from string
def balancedString(str):
    count, i = 0, 0
    n = len(str)

    # Maintain a count for opening 
    # brackets Traversing string
    for i in range(n):

        # check if opening bracket
        if (str[i] == '('):

            # print str[i] and increment 
            # count by 1
            print(str[i], end = "")
            count += 1
            
        # check if closing bracket and count != 0
        elif (str[i] == ')' and count != 0):
            print(str[i], end = "")

            # decrement count by 1
            count -= 1
            
        # if str[i] not a closing brackets
        # print it
        elif (str[i] != ')'):
            print(str[i], end = "")
            
    # balanced brackets if opening brackets
    # are more then closing brackets
    if (count != 0):
        
        # print remaining closing brackets
        for i in range(count):
            print(")", end = "")

# Driver code
if __name__ == '__main__':
    str = "gau)ra)v(ku(mar(rajput))"
    balancedString(str)

# This code is contributed by 29AjayKumar
C# 
// C# implementation of above approach
using System;

class GFG 
{ 

// Print balanced and remove 
// extra brackets from string 
public static void balancedString(String str) 
{ 
    int count = 0, i; 
    int n = str.Length; 

    // Maintain a count for opening 
    // brackets Traversing string 
    for (i = 0; i < n; i++) 
    { 

        // check if opening bracket 
        if (str[i] == '(')
        { 

            // print str[i] and increment 
            // count by 1 
            Console.Write(str[i]); 
            count++; 
        } 

        // check if closing bracket 
        // and count != 0 
        else if (str[i] == ')' && count != 0) 
        { 
            Console.Write(str[i]); 

            // decrement count by 1 
            count--; 
        } 

        // if str[i] not a closing 
        // brackets print it 
        else if (str[i] != ')') 
            Console.Write(str[i]); 
    } 

    // balanced brackets if opening 
    // brackets are more then closing 
    // brackets 
    if (count != 0) 
    
        // print remaining closing brackets 
        for (i = 0; i < count; i++) 
            Console.Write(")"); 
} 

// Driver Code
public static void Main() 
{ 
    String str = "gau)ra)v(ku(mar(rajput))"; 
    balancedString(str); 
} 
} 

// This code is contributed 
// by PrinciRaj1992
PHP 
<?php
// PHP implementation of above approach

// Print balanced and remove 
// extra brackets from string
function balancedString($str)
{
    $count = 0;
    $n = strlen($str);

    // Maintain a count for opening 
    // brackets Traversing string
    for ($i = 0; $i < $n; $i++) 
    {

        // check if opening bracket
        if ($str[$i] == '(') 
        {

            // print str[i] and increment 
            // count by 1
            echo $str[$i];
            $count++;
        }

        // check if closing bracket and count != 0
        else if ($str[$i] == ')' && $count != 0) 
        {
            echo $str[$i];

            // decrement count by 1
            $count--;
        }

        // if str[i] not a closing brackets
        // print it
        else if ($str[$i] != ')')
            echo $str[$i];
    }

    // balanced brackets if opening brackets
    // are more then closing brackets
    if ($count != 0)
    
        // print remaining closing brackets
        for ($i = 0; $i < $count; $i++)
        echo ")";
}

// Driver code
$str = "gau)ra)v(ku(mar(rajput))";
balancedString($str);

// This code is contributed
// by Akanksha Rai
?>
JavaScript 
<script>

// javascript implementation of above approach

// Print balanced and remove
// extra brackets from string
function balancedString( str) 
{ 
    var count = 0, i; 
    var n = str.length; 
  
    // Maintain a count for opening
    // brackets Traversing string
    for (i = 0; i < n; i++) 
    { 
  
        // check if opening bracket
        if (str[i] == '(')
        { 
  
            // print str[i] and increment
            // count by 1
            document.write(str[i]); 
            count++; 
        } 
  
        // check if closing bracket
        // and count != 0
        else if (str[i] == ')' && count != 0) 
        { 
            document.write(str[i]); 
  
            // decrement count by 1
            count--; 
        } 
  
        // if str[i] not a closing
        // brackets print it
        else if (str[i] != ')') 
            document.write(str[i]); 
    } 
  
    // balanced brackets if opening
    // brackets are more then closing
    // brackets
    if (count != 0) 
      
        // print remaining closing brackets
        for (i = 0; i < count; i++) 
            document.write(")"); 
} 
  
// Driver Code
 
    var str = "gau)ra)v(ku(mar(rajput))"; 
    balancedString(str); 

// This code is contributed by bunnyram19.
</script>

Output
gaurav(ku(mar(rajput)))

Complexity Analysis:

  • Time Complexity: O(N), where N is the size of the given string.
  • Auxiliary Space: O(1) as no extra space is being used.

Next Article
Balance a string after removing extra brackets

R
rajput-ji
Improve
Article Tags :
Practice Tags :

Similar Reads

We use cookies to ensure you have the best browsing experience on our website. By using our site, you acknowledge that you have read and understood our Cookie Policy & Privacy Policy
Lightbox