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 = βgau)ra)v(ku(mar(rajput))β
Output: gaurav(ku(mar(rajput)))Input: str = β1+5)+5+)6+(5+9)*9β
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 β(β then print that bracket and increment count.
- Check if the element at current index is a closing bracket β)β 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 β)β 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.