Minimum number of Parentheses to be added to make it valid

Given a string S of parentheses ‘(‘ or ‘)’ where, 0\leq len(S)\leq 1000. The task is to find minimum number of parentheses ‘(‘ or ‘)’ (at any positions) we must add to make the resulting parentheses string is valid.

Examples:

Input: str = "())"
Output: 1
One '(' is required at beginning.

Input: str = "((("
Output: 3
Three ')' is required at end.



Approach: We keep the track of balance of the string i:e the number of ‘(‘ minus the number of ‘)’. A string is valid if its balance is 0, and also every prefix has non-negative balance.

Now, consider the balance of every prefix of S. If it is ever negative (say, -1), we must add a ‘(‘ bracket at the beginning. Also, if the balance of S is positive (say, +P), we must add P times ‘)’ brackets at the end.

Below is the implementation of the above approach:

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

// C++ Program to find minimum number of '(' or ')'
// must be added to make Parentheses string valid.
#include <bits/stdc++.h>
using namespace std;
  
// Function to return required minimum number
int minParentheses(string p)
{
  
    // maintain balance of string
    int bal = 0;
    int ans = 0;
  
    for (int i = 0; i < p.length(); ++i) {
  
        bal += p[i] == '(' ? 1 : -1;
  
        // It is guaranteed bal >= -1
        if (bal == -1) {
            ans += 1;
            bal += 1;
        }
    }
  
    return bal + ans;
}
  
// Driver code
int main()
{
  
    string p = "())";
  
    // Function to print required answer
    cout << minParentheses(p);
  
    return 0;
}

chevron_right


Java

filter_none

edit
close

play_arrow

link
brightness_4
code

// Java Program to find minimum number of '(' or ')' 
// must be added to make Parentheses string valid. 
  
public class GFG {
  
    // Function to return required minimum number 
    static int minParentheses(String p) 
    
        
        // maintain balance of string 
        int bal = 0
        int ans = 0
        
        for (int i = 0; i < p.length(); ++i) { 
        
            bal += p.charAt(i) == '(' ? 1 : -1
        
            // It is guaranteed bal >= -1 
            if (bal == -1) { 
                ans += 1
                bal += 1
            
        
        
        return bal + ans; 
    
      
    public static void main(String args[])
    {
        String p = "())"
          
        // Function to print required answer 
        System.out.println(minParentheses(p)); 
        
    }
    // This code is contributed by ANKITRAI1
}

chevron_right


Python3

filter_none

edit
close

play_arrow

link
brightness_4
code

# Python3 Program to find 
# minimum number of '(' or ')' 
# must be added to make Parentheses 
# string valid. 
  
# Function to return required 
# minimum number 
def minParentheses(p):
      
    # maintain balance of string 
    bal=0
    ans=0
    for i in range(0,len(p)):
        if(p[i]=='('):
            bal+=1
        else:
            bal+=-1
              
        # It is guaranteed bal >= -1
        if(bal==-1):
            ans+=1
            bal+=1
    return bal+ans
  
# Driver code
if __name__=='__main__':
    p = "())"
      
# Function to print required answer 
    print(minParentheses(p))
      
# this code is contributed by 
# sahilshelangia

chevron_right


C#

filter_none

edit
close

play_arrow

link
brightness_4
code

// C# Program to find minimum number 
// of '(' or ')' must be added to 
// make Parentheses string valid. 
using System;
  
class GFG
{
// Function to return required 
// minimum number 
static int minParentheses(string p) 
  
    // maintain balance of string 
    int bal = 0; 
    int ans = 0; 
  
    for (int i = 0; i < p.Length; ++i) 
    
  
        bal += p[i] == '(' ? 1 : -1; 
  
        // It is guaranteed bal >= -1 
        if (bal == -1) 
        
            ans += 1; 
            bal += 1; 
        
    
  
    return bal + ans; 
  
// Driver code 
public static void Main() 
    string p = "())"
  
    // Function to print required answer 
    Console.WriteLine(minParentheses(p)); 
}
  
// This code is contributed 
// by Kirti_Mangal

chevron_right


PHP

filter_none

edit
close

play_arrow

link
brightness_4
code

<?php 
// PHP Program to find minimum number of 
// '(' or ')' must be added to make 
// Parentheses string valid.
  
// Function to return required minimum number
function minParentheses($p)
{
  
    // maintain balance of string
    $bal = 0;
    $ans = 0;
  
    for ($i = 0; $i < strlen($p); ++$i
    {
        if ($p[$i] == '(' )
            $bal += 1 ;
        else
            $bal += -1;
  
        // It is guaranteed bal >= -1
        if ($bal == -1) 
        {
            $ans += 1;
            $bal += 1;
        }
    }
  
    return $bal + $ans;
}
  
// Driver code
$p = "())";
  
// Function to print required answer
echo minParentheses($p);
  
// This code is contributed by ita_c
?>

chevron_right


Output:

1

Time Complexity: O(N), where N is the length of S.
Space Complexity: O(1).



My Personal Notes arrow_drop_up

Check out this Author's contributed articles.

If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.