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Convert an unbalanced bracket sequence to a balanced sequence

  • Difficulty Level : Hard
  • Last Updated : 13 May, 2021

Given an unbalanced bracket sequence of ‘(‘ and ‘)’, convert it into a balanced sequence by adding the minimum number of ‘(‘ at the beginning of the string and ‘)’ at the end of the string.

Examples: 

Input: str = “)))()” 
Output: “((()))()”

Input: str = “())())(())())” 
Output: “(((())())(())())”



Approach: 

  • Let us assume that whenever we encounter with opening bracket the depth increases by one and with a closing bracket the depth decreases by one.
  • Find the maximum negative depth in minDep and add that number of ‘(‘ at the beginning.
  • Then loop the new sequence to find the number of ‘)’s needed at the end of the string and add them.
  • Finally, return the string.

Below is the implementation of the approach: 

C++




// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
 
// Function to return balancedBrackets string
string balancedBrackets(string str)
{
    // Initializing dep to 0
    int dep = 0;
 
    // Stores maximum negative depth
    int minDep = 0;
 
    for (int i = 0; i < str.length(); i++)
    {
        if (str[i] == '(')
            dep++;
        else
            dep--;
 
        // if dep is less than minDep
        if (minDep > dep)
            minDep = dep;
    }
 
    // if minDep is less than 0 then there
    // is need to add '(' at the front
    if (minDep < 0)
    {
        for (int i = 0; i < abs(minDep); i++)
            str = '(' + str;
    }
 
    // Reinitializing to check the updated string
    dep = 0;
 
    for (int i = 0; i < str.length(); i++)
    {
        if (str[i] == '(')
            dep++;
        else
            dep--;
    }
 
    // if dep is not 0 then there
    // is need to add ')' at the back
    if (dep != 0)
    {
        for (int i = 0; i < dep; i++)
            str = str + ')';
    }
    return str;
}
 
// Driver code
int main()
{
    string str = ")))()";
    cout << balancedBrackets(str);
}
Java



// Java implementation of the approach
import java.util.*;
 
class GFG {
 
    // Function to return balancedBrackets string
    static String balancedBrackets(String str)
    {
        // Initializing dep to 0
        int dep = 0;
 
        // Stores maximum negative depth
        int minDep = 0;
 
        for (int i = 0; i < str.length(); i++)
        {
            if (str.charAt(i) == '(')
                dep++;
            else
                dep--;
 
            // if dep is less than minDep
            if (minDep > dep)
                minDep = dep;
        }
 
        // if minDep is less than 0 then there
        // is need to add '(' at the front
        if (minDep < 0)
        {
            for (int i = 0; i < Math.abs(minDep); i++)
                str = '(' + str;
        }
 
        // Reinitializing to check the updated string
        dep = 0;
 
        for (int i = 0; i < str.length(); i++)
        {
            if (str.charAt(i) == '(')
                dep++;
            else
                dep--;
        }
 
        // if dep is not 0 then there
        // is need to add ')' at the back
        if (dep != 0)
        {
            for (int i = 0; i < dep; i++)
                str = str + ')';
        }
        return str;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        String str = ")))()";
        System.out.println(balancedBrackets(str));
    }
}
 
// This code is contributed by ihritik
Python3



# Python3 implementation of the approach
 
# Function to return balancedBrackets String
 
 
def balancedBrackets(Str):
 
    # Initializing dep to 0
    dep = 0
 
    # Stores maximum negative depth
    minDep = 0
 
    for i in Str:
        if (i == '('):
            dep += 1
        else:
            dep -= 1
 
        # if dep is less than minDep
        if (minDep > dep):
            minDep = dep
 
    # if minDep is less than 0 then there
    # is need to add '(' at the front
    if (minDep < 0):
        for i in range(abs(minDep)):
            Str = '(' + Str
 
    # Reinitializing to check the updated String
    dep = 0
 
    for i in Str:
        if (i == '('):
            dep += 1
        else:
            dep -= 1
 
    # if dep is not 0 then there
    # is need to add ')' at the back
    if (dep != 0):
        for i in range(dep):
            Str = Str + ')'
 
    return Str
 
 
# Driver code
Str = ")))()"
print(balancedBrackets(Str))
 
# This code is contributed by Mohit Kumar
C#



// C# implementation of the approach
using System;
 
class GFG {
 
    // Function to return balancedBrackets string
    static string balancedBrackets(string str)
    {
        // Initializing dep to 0
        int dep = 0;
 
        // Stores maximum negative depth
        int minDep = 0;
 
        for (int i = 0; i < str.Length; i++)
        {
            if (str[i] == '(')
                dep++;
            else
                dep--;
 
            // if dep is less than minDep
            if (minDep > dep)
                minDep = dep;
        }
 
        // if minDep is less than 0 then there
        // is need to add '(' at the front
        if (minDep < 0)
        {
            for (int i = 0; i < Math.Abs(minDep); i++)
                str = '(' + str;
        }
 
        // Reinitializing to check the updated string
        dep = 0;
 
        for (int i = 0; i < str.Length; i++)
        {
            if (str[i] == '(')
                dep++;
            else
                dep--;
        }
 
        // if dep is not 0 then there
        // is need to add ')' at the back
        if (dep != 0)
        {
            for (int i = 0; i < dep; i++)
                str = str + ')';
        }
        return str;
    }
 
    // Driver code
    public static void Main()
    {
        String str = ")))()";
        Console.WriteLine(balancedBrackets(str));
    }
}
 
// This code is contributed by ihritik
Javascript



<script>
      // JavaScript implementation of the approach
      // Function to return balancedBrackets string
      function balancedBrackets(str) {
        // Initializing dep to 0
        var dep = 0;
 
        // Stores maximum negative depth
        var minDep = 0;
 
        for (var i = 0; i < str.length; i++) {
          if (str[i] === "(") dep++;
          else dep--;
 
          // if dep is less than minDep
          if (minDep > dep) minDep = dep;
        }
 
        // if minDep is less than 0 then there
        // is need to add '(' at the front
        if (minDep < 0) {
          for (var i = 0; i < Math.abs(minDep); i++) str = "(" + str;
        }
 
        // Reinitializing to check the updated string
        dep = 0;
 
        for (var i = 0; i < str.length; i++) {
          if (str[i] === "(") dep++;
          else dep--;
        }
 
        // if dep is not 0 then there
        // is need to add ')' at the back
        if (dep !== 0) {
          for (var i = 0; i < dep; i++) str = str + ")";
        }
        return str;
      }
 
      // Driver code
      var str = ")))()";
      document.write(balancedBrackets(str));
    </script>
Output
((()))()
Time Complexity: O(N)
Auxiliary Space: O(N)



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Pratims10
@Pratims10
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