Balance a string after removing extra brackets

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++

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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); 

chevron_right


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#

filter_none

edit
close

play_arrow

link
brightness_4
code

// 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

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?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
?>

chevron_right


Output:

gaurav(ku(mar(rajput)))


My Personal Notes arrow_drop_up

Strategy Path planning and Destination matters in success No need to worry about in between temporary failures

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.





Article Tags :
Practice Tags :


5


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.