Generate Parentheses Algorithm: Backtracking Solution in JavaScript

Understanding the Generate Parentheses Problem in JavaScript 🌀

If you’ve ever tried solving algorithmic problems, you might have encountered one that asks you to generate all combinations of valid parentheses for a given number n. But what does it mean for a combination to be valid? And how do you go about generating all possible combinations?

In this blog post, we’re going to break down the Generate Parentheses problem step by step and see how we can solve it using JavaScript. Whether you're new to coding or just learning algorithms, this post will guide you through the logic in a simple, beginner-friendly way. Let’s dive in! 🚀

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Problem Statement:

Given an integer n, your task is to generate all combinations of n pairs of valid parentheses.

Valid parentheses are those where:

• Every opening parenthesis ( has a corresponding closing parenthesis ).
• The parentheses are balanced. This means at no point in the string should there be more closing parentheses ) than opening parentheses (.

For example, if n = 3, the valid combinations would be:

["((()))", "(()())", "(())()", "()(())", "()()()"]

Now, let’s dive into how we can solve this problem using JavaScript!

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Step 1: The Backtracking Approach

To solve this problem, we’re going to use a technique called backtracking. Backtracking is a method where you try all possibilities (in this case, adding parentheses) and backtrack when you reach a dead end.

Here’s the basic idea:

• You start with an empty string for the parentheses.
• You can add an opening parenthesis ( if you haven't used all n opening parentheses yet.
• You can add a closing parenthesis ) if you’ve already added more opening parentheses than closing ones.
• If you’ve used n opening parentheses and n closing parentheses, you have a valid combination!

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Step 2: The Code

Here’s how you can implement the solution in JavaScript:

const generateParentheses = (n) => {
  const res = [] // This will store the valid combinations

  const solve = (chres, openParenthese, closedParenthese) => {
    // If we have used n opening and n closing parentheses, we have a valid combination
    if (openParenthese === n && closedParenthese === n) {
      res.push(chres) // Add the valid combination to the result
      return
    }

    // If we still have room to add an opening parenthesis
    if (openParenthese < n) {
      solve(chres + '(', openParenthese + 1, closedParenthese) // Add '(' and recurse
    }

    // If we can add a closing parenthesis (only if we have more opening parentheses)
    if (closedParenthese < openParenthese) {
      solve(chres + ')', openParenthese, closedParenthese + 1) // Add ')' and recurse
    }
  }

  solve('', 0, 0) // Call the solve function to start the backtracking

  return res // Return the result containing all valid combinations
}

export { generateParentheses }

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Step 3: How the Code Works

Let’s break down how the code works:

1. generateParentheses(n): This is the main function. It initializes an empty array res to hold the valid combinations of parentheses. Then, it calls the solve() function, which performs the backtracking.

2. solve(chres, openParenthese, closedParenthese):

   • chres: The current string being built, representing a combination of parentheses.
   • openParenthese: The count of opening parentheses ( used so far.
   • closedParenthese: The count of closing parentheses ) used so far.

3. Base Case: If we’ve used n opening parentheses and n closing parentheses, we have a valid combination, so we add it to the result array res.

4. Recursive Cases:

   • We can add an opening parenthesis ( if we haven't used all n opening parentheses yet.
   • We can add a closing parenthesis ) if we’ve used more opening parentheses than closing ones.

5. Return the Result: Once the backtracking completes, the function returns the array res, which contains all valid combinations.

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Step 4: Example Usage

Let’s see how we can use this function to generate parentheses combinations for a given number n. Here’s an example:

const n = 3
const result = generateParentheses(n)
console.log(result) // Outputs: ["((()))", "(()())", "(())()", "()(())", "()()()"]

For n = 3, the output will be:

["((()))", "(()())", "(())()", "()(())", "()()()"]

This is exactly what we expect! All combinations of 3 pairs of parentheses are valid.

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Step 5: Why This Works

By using backtracking, we systematically explore all possible combinations of parentheses. We ensure that:

• We never add more closing parentheses than opening ones.
• We only generate combinations that are valid and balanced.

This approach is both efficient and easy to understand, making it an excellent solution for beginners learning about algorithms.

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Conclusion

In this post, we’ve learned how to solve the Generate Parentheses problem using JavaScript and the backtracking technique. Backtracking is a powerful tool for solving problems that require exploring all possibilities, and we’ve applied it to generate all valid parentheses combinations.

I hope this explanation was easy to follow and helped you understand the problem and solution. Happy coding! 😄