Understanding function to generate parentheses

Question

I have this algorithm to generate all combinations of well-formed parentheses.

Can someone explain the core concept of the algorithm? I tried debugging through it, but I still can't seem to grasp the underlying concept behind the algorithm.

Additionally, any general advice for how one can come up with such an algorithm for this problem, i.e how did one even get so clever to solve it this way, or what practice one has to do to reach this stage.

Problem:

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:

“((()))”, “(()())”, “(())()”, “()(())”, “()()()”

Code:

public ArrayList<String> generateParenthesis(int n) {
    ArrayList<String> solutions = new ArrayList<String>();
    recursion(n, new String(), solutions);
    return solutions;
}

private void recursion(int n, String str, ArrayList<String> sol) {
    if(str.length() == 2 * n)
        sol.add(str);
    else {
        int left = 0;
        int right = 0;
        for(int i = 0; i < str.length(); ++i) {
            if(str.charAt(i) == '(')
                left++;
            if(str.charAt(i) == ')')
                right++;
        }
        if(left == right)
            recursion(n, str + "(", sol);
        else if(right < left) {
            if(left < n)
                recursion(n, str + "(", sol);
            recursion(n, str + ")", sol);
        }
    }
}

Kultur

possible duplicate of print all validate parentheses, how does the recursive work here? – Bastille

Answer 1

It helps me to see visually how the calls are being stacked. I added a parameter String depth to the call and printed out depth + str on each call, adding four spaces to each depth parameter for a new call. This gives us a good view of the call order.

Here's the code for it:

recursion(3, new String(), solutions, "");
//...
private static void recursion(int n, String str, ArrayList<String> sol, String depth) {
    System.out.println(depth + str);
    //...
        if(left == right)
            recursion(n, str + "(", sol, depth + "    ");
        else if(right < left) {
            if(left < n)
                recursion(n, str + "(", sol, depth + "    ");
            recursion(n, str + ")", sol, depth + "    ");
}

---

And here's what it prints out:

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

Each level of recursion adds another indent to the output. If two outputs are at the same level of indentation, then they were both called from the same level of recursion.

Here's another visual:

Note that each node is a deeper level of recursion and each time a child node comes straight down out of a parent node, it doesn't split into two recursive paths. That is, the parent node only calls recursion once.

Approachable

Answer 2

Recursion can definitely mess with your head. Here's another approach which might be easier to follow:

void generate() {
    ArrayList<String> results = new ArrayList<String>();
    generateParentheses(4, 0, new StringBuilder(), results);
    System.out.println(results);
}

void generateParentheses(final int remaining, final int openCount, final StringBuilder s, final List<String> results) {
    if (remaining == 0 && openCount == 0) {
        results.add(s.toString());
        return;
    }
    if (openCount > 0) { // we can close the open one
        s.append(")");
        generateParentheses(remaining, openCount-1, s, results);
        s.setLength(s.length()-1); // pop the last char off
    }
    if (remaining > 0) { // start a new one
        s.append("(");
        generateParentheses(remaining-1, openCount+1, s, results);
        s.setLength(s.length()-1); // pop the last char off
    }
}

Output is [()()()(), ()()(()), ()(())(), ()(()()), ()((())), (())()(), (())(()), (()())(), (()()()), (()(())), ((()))(), ((())()), ((()())), (((())))]

This goes at the problem from the other end. How do you come up with these patterns?

Start with the number of pairs (remaining).

There are only two possibilities: open or closed. An open parentheses can only be appended if there are some remaining to append. A close parentheses can only be appended if there is a corresponding opening parentheses to close.

So you just need to keep a count of how many remaining you have, and how deep into parentheses you are. Let recursion handle the rest.

Village

Answer 3

Here's my attempted breakdown of the provided algorithm:

if(str.length() == 2 * n)
        sol.add(str);

You know you're done if the string length is twice the number of pairs of parentheses. Why?

Because each pair of parentheses is 2 characters long, having n pairs of parentheses means you have 2 * n characters total (i.e. length / 2 == number of parentheses)

int left = 0;
int right = 0;
for(int i = 0; i < str.length(); ++i) {
    if(str.charAt(i) == '(')
        left++;
    if(str.charAt(i) == ')')
        right++;
}

This loops through the string character by character and does some tests and changes left and right depending on whether the particular character is a opening or closing parentheses. Based on that, can you figure out what left and right are at the end of the loop?

left and right are the number of opening/closing parentheses in the string, respectively

if(left == right)
    recursion(n, str + "(", sol);
else if(right < left) {
    if(left < n)
        recursion(n, str + "(", sol);
    recursion(n, str + ")", sol);
}

If left == right, then you call the same function, except you tack on an opening parenthesis onto the existing string. Why add a parenthesis, and why is it an opening parenthesis?

You know you're not done, because if you were you would have triggered the first if statement. Because adding a closed parenthesis would result in ill-formed parenthesis because there is no unclosed opening parenthesis (remember, left == right), the next logical step is to add an opening parenthesis.

If right < left, you know that you have at least one unclosed opening parenthesis. So you want to do another check.

if(left < n)
    recursion(n, str + "(", sol)

Why do this check? If left < n, then you know that if you close all the open parentheses you won't have enough pairs (n pairs of parentheses have n opening parentheses). So might as well add another opening parenthesis!

The last statement has an implicit else associated with it. If left is not < n, then you know adding another left parenthesis would put you over the requested number of pairs. So add a closing parenthesis and continue.

Abeyta

Answer 4

The core concept:

1)X:If a string has more closing parentheses than opening ones, adding more parentheses to the right won't make it a well-formed combination.

2)All well-formed combinations have equal number of opening and closing parentheses. Hence, each type can be found exactly n times

3)If the number of closing ones is less than the number of the opening ones, then we can always make a consequence well-formed, by adding more closing parentheses.

This algorithm builds the combination, adding new symbols to the right

       public ArrayList<String> generateParenthesis(int n) {
                ArrayList<String> solutions = new ArrayList<String>();
                recursion(n, new String(), solutions);
                return solutions;
            }

            private void recursion(int n, String str, ArrayList<String> sol) {
            //If we got a sting long enough, we return it. This means a) We generate all 
//strings only once. b)If a string of length 2*n is created, then it is correct. Other
//code should satisfy these conditions
                if(str.length() == 2 * n) 
                    sol.add(str);
                else {
                    int left = 0;
                    int right = 0;
                    for(int i = 0; i < str.length(); ++i) {
                        if(str.charAt(i) == '(')
                            left++;
                        if(str.charAt(i) == ')')
                            right++;
                    }
                    //left and right are now numbers of parentheses in the string. 
//Opening and closing respectively.
                    if(left == right)//On each step we maintain the condition X 
//that the number of closing brackets is less or equal to the number of opening.
//Therefore, is their numbers are equal we can only add opening ones
                        recursion(n, str + "(", sol);
                    else if(right < left) { // if X is maintained and we can add 
//both types
                        if(left < n)// The number of opened should be no more than n, 
//according to 2)
                            recursion(n, str + "(", sol);
                        recursion(n, str + ")", sol);//The number of closing ones is 
//limited by X by the number of opening ones, which is limited by n => the number of 
//the closed ones is limited by n => we can add them as long as it doesn't violate X
                    }

                }
            }

Gwendolyn

Answer 5

I wrote my own recursive parentheses generator in a different style. It basically builds up a String, but at each recursive call, a new String is created so that the backtracking is correct. I hope someone finds it helpful.

import java.util.ArrayList;
import java.util.List;

public class GenerateParentheses {

    // N: The max number of matching parentheses. This value does not change.
    // usedL, usedR : Number of left and right parentheses already used in 'current' string.
    // current: the current string being built.
    // depth: recursion depth, used for pretty-printing
    public static void generate(int N, int usedL, int usedR, String current, List<String> result, int depth) {

        System.out.printf("%susedL=%d, usedR=%d, current='%s'\n",
                    getIndentation(depth), usedL, usedR, current);

        if (usedL == N && usedR == N) {
            // We've used up all the available parentheses (up to N),
            // so add the current built string to the result.
            result.add(current);
            return;
        }

        if (usedL < N) {
            // Add another left parenthesis "(".
            String newCurrent = current + "(";
            generate(N, usedL + 1, usedR, newCurrent, result, depth+1);
        }

        if (usedR < N && usedL > usedR) {
            // Add another right parenthesis ")" if there are already 
            // used left parentheses.
            String newCurrent = current + ")";
            generate(N, usedL, usedR + 1, newCurrent, result, depth+1);
        }
    }

    // Utility function used for pretty-printing.
    private static String getIndentation(int depth) {
        StringBuilder sb = new StringBuilder();
        for (int i = 0; i < depth; i++) {
            sb.append("  ");
        }
        return sb.toString();
    }

    public static void main(String argv[]) {

        int N = 3;
        int usedL = 0;
        int usedR = 0;
        String current = "";
        List<String> result = new ArrayList<String>();
        int depth = 0;

        generate(N, usedL, usedR, current, result, depth);

        for (String s : result) {
            System.out.printf("%s\n", s);
        }
    }
}

Here is the output:

usedL=0, usedR=0, current=''
  usedL=1, usedR=0, current='('
    usedL=2, usedR=0, current='(('
      usedL=3, usedR=0, current='((('
        usedL=3, usedR=1, current='((()'
          usedL=3, usedR=2, current='((())'
            usedL=3, usedR=3, current='((()))'
      usedL=2, usedR=1, current='(()'
        usedL=3, usedR=1, current='(()('
          usedL=3, usedR=2, current='(()()'
            usedL=3, usedR=3, current='(()())'
        usedL=2, usedR=2, current='(())'
          usedL=3, usedR=2, current='(())('
            usedL=3, usedR=3, current='(())()'
    usedL=1, usedR=1, current='()'
      usedL=2, usedR=1, current='()('
        usedL=3, usedR=1, current='()(('
          usedL=3, usedR=2, current='()(()'
            usedL=3, usedR=3, current='()(())'
        usedL=2, usedR=2, current='()()'
          usedL=3, usedR=2, current='()()('
            usedL=3, usedR=3, current='()()()'
((()))
(()())
(())()
()(())
()()()

Murphey

Answer 6

How you get to the solution will be easier to see as you get more experience with recursion.

Idea: well-formed parentheses are always going to start with a left paren, have an equal number of left & right, and when reading left to right the following will always be true left >= right

Therefore when making a recursive solution, we'll use a simple rule: prefer to open left paren and let the nature of the recursive function unwinding take care of the rest

private void recursion(int n, String str, ArrayList<String> sol) {
    if(str.length() == 2 * n)
        sol.add(str);

We have n pairs, add the solution & return

    else {
        int left = 0;
        int right = 0;
        for(int i = 0; i < str.length(); ++i) {
            if(str.charAt(i) == '(')
                left++;
            if(str.charAt(i) == ')')
                right++;
        }

Count up the number of left & right parens

        if(left == right)
            recursion(n, str + "(", sol);

str is currently balanced & since we prefer left over right, add a left

        else if(right < left) {

This could just be an else, right will never be > left. Either way, it means that we're currently unbalanced, there is at least 1 more left than right.

            if(left < n)
                recursion(n, str + "(", sol);

Check if you can add another left, again because prefer left over right

            recursion(n, str + ")", sol);
        }
    }

Add our right paren. This will either close the paren added in the line above it, or if that wasn't executed, it will close an earlier left (remember that we're in this block because it's currently unbalanced)

Rimple

I have a hard time visualizing the second, and third strings it generates, on n = 3 ;, if you debug it, it just appears a bit weird. – Kultur

@Kultur Start with the first grouping and jump into the code from that point. All of the right parenthesis unwind and we return from recursion(n, str + "(", sol); when str is '((' (The third ( was added in the call, not added to the local str). the next line says to add a right paren & it goes from there – Rimple

Answer 7

Here's a more simpler and intuitive solution to your problem.

Again, this follows the recursion idea but it is easier to read and more efficient than the one you posted.

public void generateParantheses(int n){
    helper(n,0,0,"");
}
public void helper(int n, int open, int close, String result){

    if(result.length()==2*n) { 
    // easy enough to understand? one open and one close for each n?
        System.out.println(result);
        return;
    }
    if(open<n){
        //start off with all n open parantheses
        helper(n, open+1, close, result+"(" );
    }
    if(close<open){
        // now check if we can add closing parantheses on top of open in this condition
        helper(n, open, close+1, result+")");
    }
}

Spool

Answer 8

public static void findParenthisis(String s , int left ,int right){
        if(left==right && left == 0){
            System.out.println(s);
        }
        if(left > 0){
            findParenthisis(s+'(',left-1,right);
        }
        if(left < right){
            findParenthisis(s + ')',left,right-1);
        }
    }

Emptyhanded

Answer 9

    void generateParenthesis(int open, int close, int position, int n, char[] str) {
    /*
     * open  = open parenthesis
     * close = close parenthesis
     * position   = 2*n (length of combination of valid parenthesis
     * n     = pair of parenthesis
     * Algorithm:
     * 1.Check if position == 2*n --  Yes, print the str
     * 2.check if open is less than n 
     *      If true, add a open parenthesis into str and  call the function recursively by 
     *      incrementing open by 1 and position by 1
     * 3.check if close < open
     *      If true , add a close parenthesis and call the function recursively by 
     *      incrementing close by 1 and position by 1*/


    if(position ==str.length) {
        for(int i=0;i<str.length;i++) {
            System.out.print(str[i]);
        }
        System.out.println();
        return;
    }
    if(open < n) {
        str[position] = '(';
        generateParenthesis(open+1,close,position+1,n,str);

    }if(close< open) {
        str[position]=')';
        generateParenthesis(open,close+1,position+1,n,str);
    }

}

Crystallography

Answer 10

I think I found a very intuitive solution that gets inspiration from a similar problem. Imagine you have a grid of size n x n. You want to move from the left-down corner to the right-upper corner. Every move to the right can be interpreted as a ( and every move up as ). The simplest case would be a grid of 1x1. There are two ways to go: RU (right then up) and UR (up then right) the first one corresponds to () and the second one )(. The second case is invalid so we discard all the paths from the upper triangle of the grid. For n=4 :

The problem can be solved recursively. We should start from R=0, U=0. Once we get to R=n, U=n we are done and should add the partial_solution to the solutions. There are two cases:

1. When R<n we still can move right and increment R.
2. When U<R notice we can only go up as much as we stay under the diagonal. In this case we can still move up and increment U.

def move(R, U, n, solutions, partial_solution):

    if R == n and U == n:
        solutions.append(partial_solution)
        return

    if R < n:
        move(R + 1, U, n, solutions, partial_solution + '(')
    if U < R:
        move(R, U + 1, n, solutions, partial_solution + ')')

solutions = []
n=4
move(0, 0, n, solutions, '')
print(solutions)

Additional:

Note that this formulation also helps us to understand the number of ways we can form the parenthesis. Going from R=0, U=0 to R=n, U=n requires a string of R's and U's with the property that we should have exactly n R's and n U's. For example for n=3:

RRRUUU, RRURUU, URURUR, URRRUU, ...

It's a combinatorial problem and the number of ways is combination(2n, n). But we have to consider removing the upper triangle. This leaves us with 1/(n+1) * combination(2n, n) which is the Catalan number of n.

Dam

Answer 11

`class Solution: def generateParenthesis(self, n): # Backtracking result_list = list() stack = list()

    def create_parenthesis(left_count, right_count):
        if left_count == n and right_count == n:

            result_list.append("".join(stack))
        if left_count < n:
            print("Appending ( to stack")
            stack.append("(")
            before_spacing = (left_count + right_count) * "\t"
            print(before_spacing + "".join(stack))
            create_parenthesis(left_count+1, right_count)
            print("Popping from left_count < n")
            stack.pop()
        if left_count > right_count:
            print("Appending ) to stack.")
            stack.append(")")
            before_spacing = (left_count + right_count) * "\t"
            print(before_spacing + "".join(stack))
            create_parenthesis(left_count, right_count+1)
            print("Popping from left_count > right_count")
            stack.pop()
    create_parenthesis(0, 0)
    return result_list`

print(Solution().generateParenthesis(n=3))

Output: output

Hydrous

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Answer 12

class Solution:
    def generateParenthesis(self, n: int) -> list[str]:
        stack = []
        res = []
        def back_Move(openPar,closedPar):
            if openPar == closedPar==n:
                res.append("".join(stack))
                return 
            if openPar<n:
                stack.append("(")
                back_Move(openPar+1,closedPar)
                stack.pop()
            if closedPar<openPar:
                stack.append(")")
                back_Move(openPar,closedPar+1)
                stack.pop()
            back_Move(0,0)
            return res

Guilford

Answer 13

public class GenerateParentheses {

    public List<String> generateParenthesis(int n) {
        List<String> res = new ArrayList<>();
        Stack<Character> stack = new Stack<>();
        
        // We always start with 0 open and 0 closed
        backtrack(n, 0, 0, stack, res);
        return res;
    }

    private void backtrack(int n, int openN, int closedN, Stack<Character> stack, List<String> res) {

        /*
            A parentheses is only valid if we have the same number of open and closed parentheses.
            We also confirm that the number of open parentheses is the same as 'n' pairs expected
            Once we find a perfect complete match, we add it to the result array list.
         */
        if (openN == closedN && openN == n) {
            StringBuilder sb = new StringBuilder();
            for (char c : stack) {
                sb.append(c);
            }
            res.add(sb.toString());
            return;
        }

        /*
            If the number of open parentheses is less than the expected number of pairs?
            We add an opening parentheses and increase the number of open parenthesis by 1
         */
        if (openN < n) {
            stack.push('(');

            backtrack(n, openN + 1, closedN, stack, res);

            // We pop to clean up the stack as we are done with the backtracking.
            stack.pop();
        }

        /*
            If the number of closed parentheses is less than the number of open parentheses?
            We add a closing parentheses and increase the number of closed parenthesis by 1
         */
        if (closedN < openN) {
            stack.push(')');
            backtrack(n, openN, closedN + 1, stack, res);

            // We pop to clean up the stack as we are done with the backtracking.
            stack.pop();
        }
    }

    public static void main(String[] args) {
        GenerateParentheses generateParentheses = new GenerateParentheses();
        System.out.println(generateParentheses.generateParenthesis(3));
    }
}

Balloon

Checkout: dev.to/akhilpokle/… – Balloon