# 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

## Recommended: Please try your approach on {IDE} first, before moving on to the solution.

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 ` `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

 ` `

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

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