Balance a string after removing extra brackets - GeeksforGeeks

25 May, 2021
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++
Java
Python3
C#
PHP
Javascript
// C++ implementation of above approach#include <bits/stdc++.h>using namespace std; // Print balanced and remove// extra brackets from stringvoid 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 codeint main(){     string str = "gau)ra)v(ku(mar(rajput))";     balancedString(str);     return 0;}
// Java implementation of above approachclass GFG { // Print balanced and remove// extra brackets from stringpublic 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 Methodpublic static void main(String args[]){    String str = "gau)ra)v(ku(mar(rajput))";    balancedString(str);}}
# Python implementation of above approach # Print balanced and remove# extra brackets from stringdef 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 codeif __name__ == '__main__':    str = "gau)ra)v(ku(mar(rajput))"    balancedString(str) # This code is contributed by 29AjayKumar
// C# implementation of above approachusing System; class GFG{ // Print balanced and remove// extra brackets from stringpublic 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 Codepublic static void Main(){    String str = "gau)ra)v(ku(mar(rajput))";    balancedString(str);}} // This code is contributed// by PrinciRaj1992
<?php// PHP implementation of above approach // Print balanced and remove// extra brackets from stringfunction 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?>
<script> // javascript implementation of above approach // Print balanced and remove// extra brackets from stringfunction 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>
gaurav(ku(mar(rajput)))
 
ankit15697
princiraj1992
Akanksha_Rai
29AjayKumar
bunnyram19
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