Generate Parentheses

 

1. Clarify the problem: Before diving into solving the problem, let's clarify the requirements:

  • We need to generate all valid combinations of n pairs of parentheses.
  • A valid combination means that each opening parenthesis must have a corresponding closing parenthesis, and the parentheses should be balanced.

2. Analyze the problem: To solve this problem, we can use a recursive approach to generate all possible combinations of parentheses. At each step, we have two choices: either add an opening parenthesis or add a closing parenthesis. We need to make sure that we generate valid combinations by following the constraints mentioned above.

3. Design an algorithm: Here is the algorithm to solve the problem:

  1. Initialize an empty list to store the valid combinations.
  2. Start the recursion with an empty string, an initial count of open parentheses (0), an initial count of closed parentheses (0), and the desired number of pairs (n).
  3. In the recursive function:
    • If the length of the current string is equal to 2*n, add it to the list of valid combinations.
    • If the count of open parentheses is less than n, add an opening parenthesis and make a recursive call.
    • If the count of closed parentheses is less than the count of open parentheses, add a closing parenthesis and make a recursive call.
  4. Return the list of valid combinations.

4. Explain your approach: The approach involves using a recursive function to generate all valid combinations of parentheses. At each step, we have two choices: add an opening parenthesis or add a closing parenthesis. We make recursive calls while maintaining the count of open and closed parentheses to ensure that we generate valid combinations.

5. Write clean and readable code:

python
def generateParenthesis(n):
    def backtrack(res, curr, open_count, close_count, max_pairs):
        if len(curr) == 2 * max_pairs:
            res.append(curr)
            return

        if open_count < max_pairs:
            backtrack(res, curr + "(", open_count + 1, close_count, max_pairs)

        if close_count < open_count:
            backtrack(res, curr + ")", open_count, close_count + 1, max_pairs)

    result = []
    backtrack(result, "", 0, 0, n)
    return result

# Test case
n = 3
print(generateParenthesis(n))
# Expected output: ["((()))","(()())","(())()","()(())","()()()"]

6. Test your code: Let's test the code with the given test case:

  • Test case: n = 3.
    • The expected output is ["((()))","(()())","(())()","()(())","()()()"] because these are all the valid combinations of 3 pairs of parentheses.

7. Optimize if necessary: The current solution already has an optimal time complexity of O(4^n / sqrt(n)) since there are 2n positions in the output string, and at each position, we can have 2 choices (open or close) or 0 choices if it violates the constraints.

8. Handle error cases: The code assumes that the input n is a non-negative integer. If the input is invalid, such as a negative number or a non-integer, the code may produce unexpected results or raise an error. It's important to handle such error cases gracefully.

9. Discuss complexity analysis:

  • The time complexity of the solution is O(4^n / sqrt(n)) since we generate all possible combinations of parentheses, and at each step, we have two choices (open or close) or 0 choices if it violates the constraints.
  • The space complexity is O(4^n / sqrt(n)) as well because we need to store all the valid combinations of parentheses.
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