Generate Valid Parentheses Combinations

This problem challenges you to generate all possible and valid combinations of parentheses, given a specific number of pairs. Valid parentheses combinations follow the rule that every opening parenthesis must have a corresponding closing parenthesis, and they must be properly nested. This is a classic problem in computer science often used to illustrate backtracking and recursion.

Problem Description

You are given an integer n, which represents the number of pairs of parentheses you need to generate. Your task is to return a list of all possible valid combinations of n pairs of parentheses. A valid combination means that for every opening parenthesis '(', there is a corresponding closing parenthesis ')' later in the string, and at any point in the string, the number of closing parentheses does not exceed the number of opening parentheses.

What needs to be achieved: Generate all valid combinations of parentheses. Key requirements:

• Each combination must contain 2*n characters (n opening and n closing parentheses).
• The combinations must be valid, adhering to the nesting rules.
• The output should be a list of strings. Expected behavior: The function should return a list containing all valid combinations. The order of the combinations in the list does not matter. Edge cases to consider:
• n = 0: Should return an empty list [].
• n = 1: Should return ["()"].
• Larger values of n will require generating many combinations, so efficiency is important.

Examples

Example 1:

Input: n = 3
Output: ["((()))","(()())","(())()","()(())","()()()"]
Explanation: These are all the valid combinations of 3 pairs of parentheses.

Example 2:

Input: n = 1
Output: ["()"]
Explanation: The only valid combination of 1 pair of parentheses is "()".

Example 3:

Input: n = 0
Output: []
Explanation: When n is 0, there are no parentheses to generate, so an empty list is returned.

Constraints

• 0 <= n <= 10
• The input n is an integer.
• The output list should contain only strings consisting of '(' and ')' characters.
• The time complexity should be reasonable for n up to 10. A brute-force approach that generates all possible combinations and then filters for validity is acceptable, but more efficient solutions using backtracking are preferred.

Notes

A backtracking approach is generally the most efficient way to solve this problem. Consider maintaining a count of open and closed parentheses as you build each combination. If the count of closed parentheses ever exceeds the count of open parentheses, or if you've used all available open parentheses but still have closed parentheses to use, then the current combination is invalid and you should backtrack. Think about how to systematically explore the possible combinations while ensuring validity at each step. Recursion is a natural fit for implementing backtracking.