Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

Cost to Balance the parentheses

  • Difficulty Level : Medium
  • Last Updated : 25 May, 2021

Parenthesis are said to be balanced when every opening brace has a closing brace like “()()” or “(())” or “(()())” etc. Incorrect balancing includes “)(” or “))((”  etc. The task here is to correct the sequence of parenthesis in such a way that it is done in minimum cost. And shifting of parenthesis by over one parentheses costs 1. If the parenthesis can’t be balanced then print -1.
Examples :

Input : () 
Output : 0 
Explanation : Already balanced
Input : ))(( 
Output : 4 
Explanation : Firstly, ) at position 1st goes to the last position, costing 3, so we get )((). Then, ) at position 1st goes to the 2nd position costing 1. So, finally we get ()(). Therefore, the total cost is 4.

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the DSA Self Paced Course at a student-friendly price and become industry ready.  To complete your preparation from learning a language to DS Algo and many more,  please refer Complete Interview Preparation Course.

In case you wish to attend live classes with experts, please refer DSA Live Classes for Working Professionals and Competitive Programming Live for Students.

Algorithm :  



  1. Store the braces in string.
  2. Run a loop to string size to store the count of opening and closing braces.
  3. Check if number of opening brace is equal to number of closing brace or not.
  4. If the braces are not equal then print -1 depicting that the string cant be balanced. Else proceed further.
  5. Initially, Check at 0th index that whether the string contains opening brace or closing brace. If we get an opening brace then store +1 in the array at index 0, else if closing brace is present then place -1 at 0th index.
  6. Now run a loop from 1st index to array length.
    • If opening brace is present at index i then add +1 to value at previous index i.e. i-1 and store sum at index i.
    • If closing brace is present at index i then add -1 to value at previous index i.e. i-1 and store sum at index i.
    • If value at index i is negative i.e less than 0, then add the absolute value of array[i] into a variable(ans in below program).
  7. Finally we get the minimum cost in variable ans.

Below is the implementation of above approach :  

C++




// CPP code to calculate the minimum cost
// to make the given parentheses balanced
#include <bits/stdc++.h>
using namespace std;
 
int costToBalance(string s)
{
    if (s.length() == 0)
        cout << 0 << endl;
 
    // To store absolute count of
    // balanced and unbalanced parenthesis
    int ans = 0;
 
    // o(open bracket) stores count of '(' and
    // c(close bracket) stores count of ')'
    int o = 0, c = 0;
 
    for (int i = 0; i < s.length(); i++) {
        if (s[i] == '(')
            o++;
        if (s[i] == ')')
            c++;
    }
 
    if (o != c)
        return -1;
 
    int a[s.size()];
    if (s[0] == '(')
        a[0] = 1;
    else
        a[0] = -1;
 
    if (a[0] < 0)
        ans += abs(a[0]);
 
    for (int i = 1; i < s.length(); i++) {
        if (s[i] == '(')
            a[i] = a[i - 1] + 1;
        else
            a[i] = a[i - 1] - 1;
        if (a[i] < 0)
            ans += abs(a[i]);
    }
 
    return ans;
}
 
// Driver code
int main()
{
    string s;
    s = ")))(((";
 
    cout << costToBalance(s) << endl;
 
    s = "))((";
    cout << costToBalance(s) << endl;
 
    return 0;
}

Java




// Java code to calculate the
// minimum cost to make the
// given parentheses balanced
import java.io.*;
 
class GFG
{
    static int costToBalance(String s)
    {
        if (s.length() == 0)
            System.out.println(0);
     
        // To store absolute count
        // of balanced and unbalanced
        // parenthesis
        int ans = 0;
     
        // o(open bracket) stores count
        // of '(' and c(close bracket)
        // stores count of ')'
        int o = 0, c = 0;
     
        for (int i = 0; i < s.length(); i++)
        {
            if (s.charAt(i) == '(')
                o++;
            if (s.charAt(i) == ')')
                c++;
        }
     
        if (o != c)
            return -1;
     
        int []a = new int[s.length()];
        if (s.charAt(0) == '(')
            a[0] = 1;
        else
            a[0] = -1;
     
        if (a[0] < 0)
            ans += Math.abs(a[0]);
     
        for (int i = 1; i < s.length(); i++)
        {
            if (s.charAt(i) == '(')
                a[i] = a[i - 1] + 1;
            else
                a[i] = a[i - 1] - 1;
            if (a[i] < 0)
                ans += Math.abs(a[i]);
        }
     
        return ans;
    }
     
    // Driver code
    public static void main(String args[])
    {
        String s;
        s = ")))(((";
     
        System.out.println(costToBalance(s));
     
        s = "))((";
        System.out.println(costToBalance(s));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Python3




# Python 3 code to calculate the minimum cost
# to make the given parentheses balanced
def costToBalance(s):
    if (len(s) == 0):
        print(0)
 
    # To store absolute count of
    # balanced and unbalanced parenthesis
    ans = 0
 
    # o(open bracket) stores count of '(' and
    # c(close bracket) stores count of ')'
    o = 0
    c = 0
 
    for i in range(len(s)):
        if (s[i] == '('):
            o += 1
        if (s[i] == ')'):
            c += 1
 
    if (o != c):
        return -1
 
    a = [0 for i in range(len(s))]
    if (s[0] == '('):
        a[0] = 1
    else:
        a[0] = -1
 
    if (a[0] < 0):
        ans += abs(a[0])
 
    for i in range(1, len(s)):
        if (s[i] == '('):
            a[i] = a[i - 1] + 1
        else:
            a[i] = a[i - 1] - 1
        if (a[i] < 0):
            ans += abs(a[i])
 
    return ans
 
# Driver code
if __name__ == '__main__':
    s = ")))((("
 
    print(costToBalance(s))
    s = "))(("
    print(costToBalance(s))
     
# This code is contributed by
# Surendra_Gangwar

C#




// C# code to calculate the
// minimum cost to make the
// given parentheses balanced
using System;
 
class GFG
{
    static int costToBalance(string s)
    {
        if (s.Length == 0)
            Console.WriteLine(0);
     
        // To store absolute count
        // of balanced and unbalanced
        // parenthesis
        int ans = 0;
     
        // o(open bracket) stores count
        // of '(' and c(close bracket)
        // stores count of ')'
        int o = 0, c = 0;
     
        for (int i = 0; i < s.Length; i++)
        {
            if (s[i] == '(')
                o++;
            if (s[i] == ')')
                c++;
        }
     
        if (o != c)
            return -1;
     
        int []a = new int[s.Length];
        if (s[0] == '(')
            a[0] = 1;
        else
            a[0] = -1;
     
        if (a[0] < 0)
            ans += Math.Abs(a[0]);
     
        for (int i = 1; i < s.Length; i++)
        {
            if (s[i] == '(')
                a[i] = a[i - 1] + 1;
            else
                a[i] = a[i - 1] - 1;
            if (a[i] < 0)
                ans += Math.Abs(a[i]);
        }
     
        return ans;
    }
     
    // Driver code
    static void Main()
    {
        string s;
        s = ")))(((";
     
        Console.WriteLine (costToBalance(s));
     
        s = "))((";
        Console.WriteLine (costToBalance(s));
    }
}
 
// This code is contributed by
// Manish Shaw(manishshaw1)

Javascript




<script>
 
// javascript code to calculate the
// minimum cost to make the
// given parentheses balanced
 
    function costToBalance( s)
    {
        if (s.length == 0)
            document.write(0);
     
        // To store absolute count
        // of balanced and unbalanced
        // parenthesis
        var ans = 0;
     
        // o(open bracket) stores count
        // of '(' and c(close bracket)
        // stores count of ')'
        var o = 0, c = 0;
     
        for (var i = 0; i < s.length; i++)
        {
            if (s[i] == '(')
                o++;
            if (s[i] == ')')
                c++;
        }
     
        if (o != c)
            return -1;
     
        var a = new Array(s.Length);
        if (s[0] == '(')
            a[0] = 1;
        else
            a[0] = -1;
     
        if (a[0] < 0)
            ans += Math.abs(a[0]);
     
        for (var i = 1; i < s.length; i++)
        {
            if (s[i] == '(')
                a[i] = a[i - 1] + 1;
            else
                a[i] = a[i - 1] - 1;
            if (a[i] < 0)
                ans += Math.abs(a[i]);
        }
     
        return ans;
    }
     
    // Driver code
        var s;
        s = ")))(((";
     
        document.write(costToBalance(s) + "<br>");
     
        s = "))((";
        document.write(costToBalance(s) + "<br>" );
 
// This code is contributed by bunnyram19.
</script>
Output: 
9
4

 

Time Complexity:O(N),N=String Length

Auxiliary Space:O(N)




My Personal Notes arrow_drop_up
Recommended Articles
Page :

Start Your Coding Journey Now!