Generate balanced parentheses in java

Question

The question is: Given n pairs of parentheses, write a function to generate all combinations of well-formed parentheses.

For example, given n = 3, a solution set is:

"((()))", "(()())", "(())()", "()(())", "()()()"

I used to solve this problem using string as following codes:

public class Solution {
public List<String> generateParenthesis(int n) {

    ArrayList<String> result = new ArrayList<String>();
    //StringBuilder s = new StringBuilder();
    generate(n, 0, 0, "", result);
    return result;
}
public void generate(int n, int left, int right, String s, ArrayList<String> result){

    //left is num of '(' and right is num of ')'
    if(left < right){
        return;
    }
    if(left == n && right == n){
        result.add(s);
        return;
    }
    if(left == n){
        //add ')' only.
        generate(n, left, right + 1, s + ")", result);
        return;
    }
    generate(n, left + 1, right, s + "(", result);
    generate(n, left, right + 1, s + ")", result);
    }
}

Now I want to solve this problem using StringBuilder, the code is like this:

import java.util.ArrayList;

public class generateParentheses {
public static ArrayList<String> generateParenthesis(int n) {

    ArrayList<String> result = new ArrayList<String>();
    StringBuilder sb = new StringBuilder();

    generate(n, 0, 0, result, sb);
    return result;
}

public static void generate(int n, int left, int right, ArrayList<String> result,
        StringBuilder sb) {

    if (left < right) {
        return;
    }
    if (left == n && right == n) {
        result.add(sb.toString());
        sb = new StringBuilder();
        return;
    }
    if (left == n) {
        generate(n, left, right + 1, result, sb.append(')'));
        return;
    }
    generate(n, left + 1, right, result, sb.append('('));
    //sb.delete(sb.length(), sb.length() + 1);
    generate(n, left, right + 1, result, sb.append(')'));
    //sb.delete(sb.length(), sb.length() + 1);
}

public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println(generateParenthesis(4));
    }
}

The result is not what I want: (((()))), (((()))))())), (((()))))())))()), (((()))))())))()))(), (((()))))())))()))()))(())), (((()))))())))()))()))(())))()).........

Is there anyone tell me what is the problem? Thank you very much.

Answer 1

You ware close. Your mistakes ware:

• trying to reset sb instead of removing only its last character

• way you want to reset sb :

  • by using sb = new StringBuilder(); you are reassigning sb which is local variable of current method, not variable of method which invoked it (Java is not pass-by-reference but pass-by-value).

  • your almost correct attempt ware commented sb.delete(sb.length(), sb.length() + 1); but here you are actually trying to remove characters starting from position sb.length() , but just like arrays indexes of character in StringBuilder are from 0 till sb.length() - 1 so is trying to remove one character after last character which effectively can't remove anything.

    What you needed here is

    sb.delete(sb.length() - 1, sb.length());

    or more readable

    sb.deleteCharAt(sb.length() - 1);

    but probably best approach in terms of performance setLength (described at bottom of answer)

     sb.setLength(sb.length() - 1);

• your logic of when to remove characters from StringBuilder is also flawed

  • you are doing it only in one place which ends (backtracks) recursive calls: after finding correct results. But what about other cases like if (left < right) or most importantly, if method will end normally like

    generate(3, 1, 1, ')'); generate(3, 1, 2, ')');//here we stop and backtrack

    Here generate(3, 1, 2, ')'); ends and removes last character from sb , but shouldn't previous method generate(3, 1, 1, ')') also remove its own ) added to StringBuilder?

  In other words you shouldn't remove last character only at end of successful condition in recursive call, but after each recursive call, to make sure that method will also remove character it adds.

So change your code to something like

public static void generate(int n, int left, int right, ArrayList<String> result,
        StringBuilder sb) {

    if (left < right) {
        return;
    }
    if (left == n && right == n) {
        result.add(sb.toString());
        return;
    }
    if (left == n) {
        generate(n, left, right + 1, result, sb.append(')'));
        sb.deleteCharAt(sb.length() - 1);// <--
        return;
    }
    generate(n, left + 1, right, result, sb.append('('));
    sb.deleteCharAt(sb.length() - 1);// <--
    generate(n, left, right + 1, result, sb.append(')'));
    sb.deleteCharAt(sb.length() - 1);// <--
}

or try writing something probably more readable like

public static void generate(int maxLength, int left, int right,
        ArrayList<String> result, StringBuilder sb) {
    if (left + right == maxLength) {
        if (left == right)
            result.add(sb.toString());
    } else if (left >= right) {
        generate(maxLength, left + 1, right, result, sb.append('('));
        sb.deleteCharAt(sb.length() - 1);

        generate(maxLength, left, right + 1, result, sb.append(')'));
        sb.deleteCharAt(sb.length() - 1);
    }
}

but while invoking you would need to set maxLength as 2*n since it is the max length StringBuilder should contain, so you would also have to change generateParenthesis(int n) to:

public static ArrayList<String> generateParenthesis(int n) {

    ArrayList<String> result = new ArrayList<String>();
    StringBuilder sb = new StringBuilder(2 * n);

    generate(2 * n, 0, 0, result, sb);
    //       ^^^^^
    return result;
}

---

Farther improvement:

If you are aiming for performance then you probably don't want to use delete or deleteCharAt because each time it creates new array and fills it with copy of values from without ones you don't want.

Instead you can use setLength method. If you will pass value which is smaller than number of currently stored characters it will set count to value passed in this method, which will effectively make characters after them irrelevant. In other words this characters will be no longer used in for instance toString() even if they will be still in StringBuilder buffer array.

Example:

StringBuilder sb = new StringBuilder("abc");  // 1
sb.setLength(2);                              // 2
System.out.println(sb);                       // 3
sb.append('d');                               // 4
System.out.println(sb);                       // 5

• In line 1 StringBuilder will allocate array for at least 3 characters (by default it uses str.length() + 16 to determine size of buffered array of characters it will store for now) and will place there characters from passed String, so it will contain

  ['a', 'b', 'c', '\\0', '\\0', ... , '\\0'] ^^^ - it will put next character here

  Index of position where next character should be placed is stored in count field and for now it is equal to 3 .

• In line 2 value of count will be set to 2 , but our array will not be changed so it will still look like

  ['a', 'b', 'c', '\\0', '\\0', ... , '\\0'] ^^^ - it will put next character here

• In line 3 new String will be created and printed, but it will be filled only with characters placed before index stored in count , which means that it will contain only a and b (array will still be unchanged).

• In line 4 you will add new character to buffer and it will be placed after "important" characters. Since number of important characters is stored in count field (and they are placed at beginning of array), next irrelevant character must be at position pointed by count , which means d will be placed at position with index 2 which means now array will look like

  ['a', 'b', 'd', '\\0', '\\0', ... , '\\0'] ^^^ - it will put next character here

  and value of count will be incremented (we added only one character so count will now become 3 ).

• In line 5 we will create and print string containing first 3 characters from array used by StringBuilder so we will see abd .

Answer 2

After carefully debuting this program, I found out the problem. The correct code is listed as follows:

import java.util.ArrayList;

public class generateParentheses {
    public static ArrayList<String> generateParenthesis(int n) {

    ArrayList<String> result = new ArrayList<String>();
    StringBuilder sb = new StringBuilder();

    generate(n, 0, 0, result, sb);
    return result;
}

public static void generate(int n, int left, int right, ArrayList<String> result,
        StringBuilder sb) {

    if (left < right) {
        return;
    }
    if (left == n && right == n) {
        result.add(sb.toString());
        //sb.delete(0,sb.length());
        return;
    }
    if (left == n) {
        generate(n, left, right + 1, result, sb.append(')'));
        //delete current ')'.
        sb.delete(sb.length() - 1, sb.length());
        return;
    }

    generate(n, left + 1, right, result, sb.append('('));
    //delete current '(' after you finish using it for next recursion.
    sb.delete(sb.length() - 1, sb.length());
    generate(n, left, right + 1, result, sb.append(')'));
    //same as above here.
    sb.delete(sb.length() - 1, sb.length());
}

public static void main(String[] args) {
    // TODO Auto-generated method stub
    System.out.println(generateParenthesis(4));


    }

}

And the result is: (((()))), ((()())), ((())()), ((()))(), (()(())), (()()()), (()())(), (())(()), (())()(), ()((())), ()(()()), ()(())(), ()()(()), ()()()()

Answer 3

The combination of a parameter, assigment and return:

..., StringBuilder sb,...
    //...
    sb = new StringBuilder();
    return sb;

does not make sense, because the parameter is passed in by value, ie, sb is a local variable the change of which has no effect in the calling environment.

If you want to clear the StringBuilder, there is method sb.delete( start, end ) that will truly affect the object referenced via sb.

Answer 4

Another much simpler way of doing this. Here, i am trying for a recursive solution where the recursive function adds balanced parentheses in 3 ways: "()"+result and result+"()" and "("+result+")" for each item in HashSet returned by the function it calls then put them in a HashSet to remove duplicates.

     import java.util.HashSet;
     import java.util.Set;

     public class BalancedParanthesis {
       public static void main(String args[]){
         int noOfBrackets = 3;
         HashSet<String> hs=new HashSet(generate(noOfBrackets));
         System.out.println(hs);
         }
         public static HashSet<String> generate(int in)
         {
          HashSet<String> hs= new HashSet<String>();
          if(in ==1)
         {
        hs.add("()");
        return hs;
         }
        else{
           Set<String> ab=generate(in-1);
           for(String each:ab)
            {
            hs.add("("+each+")");
            hs.add("()"+each);
            hs.add(each+"()");
            }
        return hs;
         }
        }
       }

Answer 5

Your code is inefficient. Here is a better approach. We make an effective brute force over the 2^N possible bracket permutations with heavy pruning(we quit the recursion immediately if there is no possible valid solution for the current parameters). Here's the code in C++:

#include <iostream>
#include <vector>
#include <string>
using namespace std;
string result;
vector<string> solutions;
int N = 10;
void generateBrackets(int pos, int balance)
{
    if(balance > N-pos) return;
    if(pos == N)
    {
        //we have a valid solution
        //generate substring from 0 to N-1
        //and push it to the vector
        string currentSolution;
        for(int i = 0; i < N; ++i)
        {
            currentSolution.push_back(result[i]);
        }
        solutions.push_back(currentSolution);
        return;
    }
    result[pos] = '(';
    generateBrackets(pos+1, balance+1);
    if(balance > 0)
    {
        result[pos] = ')';
        generateBrackets(pos+1, balance-1);
    }
}
int main()
{
    result.assign(N, 'a');
    generateBrackets(0, 0);
    cout<<"Printing solutions:\n";
    for(int i = 0; i < solutions.size(); ++i)
    {
        cout<<solutions[i]<<endl;
    }
    return 0;
}

Some clarifications: pos marks the current position in the bracket solution we are generating. If we reach position N (that means the solution is valid) we have a valid solution in the string result variable and we simple push it in the vector with the valid solutions. Now for what we use balance you may ask. We make the observation that in order a bracket permutation to be valid, at any time at any given position the count of the '(' from the start must be greater than the count of the ')'s and with balance we measure their difference, so me put a ')' at a given position only if the balance > 0.