Generate Parentheses in Python

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

A parentheses string is well-formed if and only if:

It is the empty string, or
It can be written as AB (A concatenated with B), where A and B are well-formed, or
It can be written as (A), where A is well-formed.

Example 1:

Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]

Example 2:

Input: n = 1
Output: ["()"]

Note: 1 <= n <= 8

Understanding the Problem

The core challenge of this problem is to generate all possible combinations of well-formed parentheses given n pairs. This problem is significant in various applications such as validating expressions, parsing, and compiler design. A common pitfall is to generate invalid combinations or miss some valid ones.

Approach

To solve this problem, we can use a backtracking approach. The idea is to build the string step by step, ensuring at each step that the string remains valid. We can start with an empty string and add either an opening or closing parenthesis, ensuring that the number of closing parentheses never exceeds the number of opening ones.

Naive Solution

A naive solution would be to generate all possible strings of length 2n and then filter out the valid ones. However, this approach is not optimal as it generates many invalid strings and has a high time complexity.

Optimized Solution

The optimized solution uses backtracking to generate only valid strings. This approach is more efficient as it prunes invalid paths early.

Algorithm

Here is a step-by-step breakdown of the backtracking algorithm:

Start with an empty string and two counters: open_count and close_count.
At each step, add an opening parenthesis if open_count is less than n.
Add a closing parenthesis if close_count is less than open_count.
Continue this process until the length of the string is 2n.
When the length of the string is 2n, add it to the result list.

Code Implementation

def generate_parentheses(n):
    def backtrack(s, open_count, close_count):
        # If the current string s has reached the maximum length, add it to the result
        if len(s) == 2 * n:
            result.append(s)
            return
        # If we can add an opening parenthesis, do so
        if open_count < n:
            backtrack(s + '(', open_count + 1, close_count)
        # If we can add a closing parenthesis, do so
        if close_count < open_count:
            backtrack(s + ')', open_count, close_count + 1)

    result = []
    backtrack('', 0, 0)
    return result

# Example usage
print(generate_parentheses(3))  # Output: ["((()))","(()())","(())()","()(())","()()()"]

Complexity Analysis

The time complexity of this approach is O(4^n / sqrt(n)), which is derived from the Catalan number. The space complexity is O(n) due to the recursion stack.

Edge Cases

Some potential edge cases include:

n = 0: The output should be an empty list.
n = 1: The output should be ["()"].

These cases are handled naturally by the algorithm.

Testing

To test the solution comprehensively, consider the following test cases:

Simple cases like n = 1 and n = 2.
Edge cases like n = 0.
Larger values of n to ensure performance.

Using a testing framework like unittest in Python can help automate these tests.

Thinking and Problem-Solving Tips

When approaching such problems, consider the following tips:

Break down the problem into smaller subproblems.
Use recursion and backtracking to explore all possible solutions.
Prune invalid paths early to improve efficiency.

Practice solving similar problems to improve your problem-solving skills.

Conclusion

Generating well-formed parentheses is a classic problem that helps in understanding recursion and backtracking. By practicing such problems, you can improve your algorithmic thinking and problem-solving skills.

Additional Resources

For further reading and practice, consider the following resources:

Generate Parentheses in Python (Time Complexity: O(4^n / sqrt(n)))