# Minimum number of Parentheses to be added to make it valid

Given a string S of parentheses ‘(‘ or ‘)’ where, . The task is to find a 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 1: Iterative 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++

 // C++ Program to find minimum number of '(' or ')'// must be added to make Parentheses string valid.#include using namespace std; // Function to return required minimum numberint 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 codeint main(){     string p = "())";     // Function to print required answer    cout << minParentheses(p);     return 0;}

## Java

 // 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}

## Python3

 # 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 codeif __name__=='__main__':    p = "())"     # Function to print required answer     print(minParentheses(p))     # this code is contributed by # sahilshelangia

## C#

 // 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

## PHP

 = -1        if ($bal == -1)  { $ans += 1;            $bal += 1; } }  return $bal + $ans;} // Driver code$p = "())"; // Function to print required answerecho minParentheses(\$p); // This code is contributed by ita_c?>

## Javascript

 

Output
1

Complexity Analysis:

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

Approach 2: Using Stack

We can also solve this problem using a stack data structure.

1. We push the index of every ‘(‘ character onto the stack, and whenever we encounter a ‘)’ character, we pop the top index from the stack.
2. This indicates that the corresponding ‘(‘ has been paired with the current ‘)’.
3. If at any point the stack is empty and we encounter a ‘)’ character, it means we need to add a ‘(‘ character to the beginning of the string to make it valid.
4. Similarly, if after processing the entire string there are still indices left in the stack, it means we need to add ‘)’ characters to the end of the string to make it valid.

Below is the implementation of the above approach:

## C++

 // C++ Program to find minimum number of '(' or ')'// must be added to make Parentheses string valid.#include using namespace std; // Function to return required minimum numberint minParentheses(string p){    stack<int> stk;    int ans = 0;     for (int i = 0; i < p.length(); ++i) {         if (p[i] == '(') {            stk.push(i);        }        else {            if (!stk.empty()) {                stk.pop();            }            else {                ans += 1;            }        }    }     // add remaining '(' characters to end    ans += stk.size();     return ans;} // Driver codeint main(){    string p = "())";     // Function to print required answer    cout << minParentheses(p);     return 0;}// This code is contributed by user_dtewbxkn77n

## Java

 import java.util.Stack; public class Main {    // Function to return required minimum number    public static int minParentheses(String p) {        Stack stk = new Stack<>();        int ans = 0;         for (int i = 0; i < p.length(); ++i) {            if (p.charAt(i) == '(') {                stk.push(i);            } else {                if (!stk.empty()) {                    stk.pop();                } else {                    ans += 1;                }            }        }         // add remaining '(' characters to end        ans += stk.size();         return ans;    }     // Driver code    public static void main(String[] args) {        String p = "())";         // Function to print required answer        System.out.println(minParentheses(p));    }}

## Python3

 def minParentheses(p):    stk = []    ans = 0     for i in range(len(p)):        if p[i] == '(':            stk.append(i)        else:            if len(stk) > 0:                stk.pop()            else:                ans += 1     # add remaining '(' characters to end    ans += len(stk)     return ans # Driver codep = "())"# Function to print required answerprint(minParentheses(p))

## C#

 using System;using System.Collections.Generic; public class Program {    // Function to return required minimum number    public static int MinParentheses(string p) {        Stack<int> stk = new Stack<int>();        int ans = 0;         for (int i = 0; i < p.Length; ++i) {            if (p[i] == '(') {                stk.Push(i);            } else {                if (stk.Count > 0) {                    stk.Pop();                } else {                    ans += 1;                }            }        }         // add remaining '(' characters to end        ans += stk.Count;         return ans;    }     // Driver code    public static void Main() {        string p = "())";         // Function to print required answer        Console.WriteLine(MinParentheses(p));    }}

## Javascript

 // Function to return required minimum numberfunction minParentheses(p) {  let stk = [];  let ans = 0;   for (let i = 0; i < p.length; i++) {    if (p[i] === '(') {      stk.push(i);    } else {      if (stk.length > 0) {        stk.pop();      } else {        ans += 1;      }    }  }   // add remaining '(' characters to end  ans += stk.length;   return ans;} // Driver codelet p = "())"; // Function to print required answerconsole.log(minParentheses(p));

Output
1

Time Complexity: O(n), where n is the length of the input string.
Space Complexity:  O(n)

Approach 3: Using string manipulation

• Continuously search for the substring “()” in the input string.
• If the substring “()” is found, replace it with an empty string.
• Repeat steps 1 and 2 until the substring “()” can no longer be found.
• The length of the resulting string is the minimum number of parentheses needed to make the string valid.

Here is the implementation of above approach:

## C++

 #include #include  using namespace std; int minParentheses(string s) {    while (s.find("()") != -1) {        s.replace(s.find("()"), 2, "");    }    return s.length();} int main() {    string s = "(((";    cout << minParentheses(s) << endl;    return 0;}

## Java

 public class Solution {    public static int minParentheses(String s) {        while (s.indexOf("()") != -1) {            s = s.replace("()", "");        }        return s.length();    }         public static void main(String[] args) {        String s = "(((";        System.out.println(minParentheses(s));    }}

## Python3

 def minParentheses(s: str) -> int:    while s.find("()") != -1:        s = s.replace("()", "")    return len(s) # Driver codes = "((("print(minParentheses(s))

## C#

 using System; public class Program {    public static void Main() {        string s = "(((";        Console.WriteLine(minParentheses(s));    }     public static int minParentheses(string s) {        while (s.IndexOf("()") != -1) {            s = s.Replace("()", "");        }        return s.Length;    }}

## Javascript

 function minParentheses(s) {    while(s.indexOf("()") != -1) {        s = s.replace("()", "");    }    return s.length;} // Driver codelet s = "(((";console.log(minParentheses(s));

Output
3

Time Complexity: O(n^2) where n is the length of the input string. This is because the find() and replace() operations are called repeatedly in a loop.

Auxiliary Space: O(n) where n is the length of the input string. This is because the string s is being replaced in place and no additional data structures are being used.

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