Problem:
Score : 400 points

Problem Statement

You are given a string S of length N consisting of ( and ). Your task is to insert some number of ( and ) into S to obtain a correct bracket sequence.

Here, a correct bracket sequence is defined as follows:

() is a correct bracket sequence.

If X is a correct bracket sequence, the concatenation of (, X and ) in this order is also a correct bracket sequence.

If X and Y are correct bracket sequences, the concatenation of X and Y in this order is also a correct bracket sequence.

Every correct bracket sequence can be derived from the rules above.

Find the shortest correct bracket sequence that can be obtained. If there is more than one such sequence, find the lexicographically smallest one.

Constraints

The length of S is N.

1 ≤ N ≤ 100

S consists of ( and ).

Input

Input is given from Standard Input in the following format:

N
S

Output

Print the lexicographically smallest string among the shortest correct bracket sequences that can be obtained by inserting some number of ( and ) into S.

Sample Input 1

3
())

Sample Output 1

(())

Sample Input 2

6
)))())

Sample Output 2

(((()))())

Sample Input 3

8
))))((((

Sample Output 3

(((())))(((())))

Solution in Java (OpenJDK 1.8.0) (judge verdict: Accepted)


import java.math.BigInteger;
import java.util.*;
public class Main {



public static void main(String[] args)
{
Scanner sc=new Scanner(System.in);
  int n=sc.nextInt();
  String s=sc.next();
  int ob=0;
  int cb=0;
  boolean check=false;
  Deque<Character> dq=new LinkedList<>();
  for(int i=0;i<n;i++) {
    dq.addLast(s.charAt(i));
  }
  
  Stack<Character> st=new Stack<>();
  for(int i=0;i<n;i++) {
   if(st.isEmpty()) {
     st.push(s.charAt(i));
     continue;
   }
   if(st.peek()=='('&&s.charAt(i)==')') {
     st.pop();
     continue;
   }
   st.push(s.charAt(i));
  
  }
   while(!st.isEmpty()) {
     if(st.peek()=='(') {
       dq.addLast(')');
       st.pop();
     }else {
       dq.addFirst('(');
       st.pop();
     }
   }
   for(char i:dq) {
     System.out.print(i);
   }
   System.out.println();


}
}
class Solution{

  public static int minSum(int arr[], int n, int k) 
  { 
          int dp[][][]=new int[n+1][k+1][100001];
          for(int i=0;i<=n;i++){
              for(int j=0;j<=k;j++){
                for(int kk=0;kk<100001;kk++)
                 dp[i][j][kk]=-1;
          }
          }
          int ans=solve(arr,n,n,k,dp,100000);
          return ans;
          
  } 
  static int solve(int arr[],int nn,int n,int k,int dp[][][],int prev){
     
 
    if(k==1){
          int ans=100001;

         for(int i=0;i<n;i++){
             ans=Math.min(ans,arr[i]);
         }
        
         return ans>prev?100001:ans;
  
      }
      if(n==0){
          return 100001;
      }
      if(dp[n][k][prev]!=-1) {
        return dp[n][k][prev];
      }
     int ans=100001;
     if(arr[n-1]<=prev){
     
     ans=Math.min(ans, Math.min(solve(arr,nn,n-1,k-1,dp,arr[n-1])+arr[n-1],solve(arr,nn,n-1,k,dp,prev)));

     }else{
     ans=Math.min(ans, solve(arr,nn,n-1,k,dp,prev));
     }
     
return dp[n][k][prev]=ans;
  }
  
 }