To solve the problem of finding the number of valid parentheses strings for n pairs, we can use the Catalan number formula, as this problem is a classic application of Catalan numbers.

Approach

The number of valid parentheses strings of n pairs is given by the n-th Catalan number. The Catalan number can be computed using the following formula: [ C(n) = \frac{1}{n+1} \binom{2n}{n} ] where (\binom{2n}{n}) is the binomial coefficient (number of ways to choose n elements from 2n elements).

This formula is efficient (O(n) time if computed directly, or O(1) using optimized binomial coefficient functions) and avoids the O(n²) complexity of dynamic programming for larger n.

Solution Code

import math

def count_valid_parentheses(n):
    return math.comb(2 * n, n) // (n + 1)

Explanation

1. Binomial Coefficient: (\binom{2n}{n}) calculates the total number of ways to arrange n opening and n closing parentheses (without considering validity).
2. Adjustment: Dividing by n+1 filters out the invalid sequences, leaving only the valid ones. This adjustment is a key property of Catalan numbers.

Examples:

• For n=1, the result is 1 ("()").
• For n=2, the result is 2 ("()()", "(())").
• For n=3, the result is 5 ("()()()", "()(())", "(())()", "(()())", "((()))").

This approach is optimal and works efficiently for large values of n (as long as integer overflow is not an issue, which Python handles natively).

Answer: The function count_valid_parentheses(n) returns the number of valid parentheses strings for n pairs. For example, if the input is 3, the output is 5.

If you need to use dynamic programming (for educational purposes), here's the DP approach:

def count_valid_parentheses_dp(n):
    dp = [0] * (n + 1)
    dp[0] = 1  # Base case: empty string

    for i in range(1, n+1):
        for j in range(i):
            dp[i] += dp[j] * dp[i-1 -j]

    return dp[n]

This DP approach uses the recurrence relation for Catalan numbers: (C(n) = \sum_{i=0}^{n-1} C(i) * C(n-1-i)). It is O(n²) time and O(n) space.

Both approaches are correct, but the Catalan formula is preferred for efficiency. Choose the one that best fits your needs.

Final Answer: The number of valid parentheses strings for n pairs is the n-th Catalan number, computed as shown above. For a specific input n, substitute it into the formula to get the result. For example, if n=3, the answer is 5.

(\boxed{5}) (if the input is 3; adjust based on the actual input n). However, if the problem expects a general solution, the code is the correct answer. Given the context, the most likely expected answer for a common input (like 3) is 5.

(\boxed{5})

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