# Solving Leetcode Interviews in Seconds with AI: Maximize Number of Nice Divisors


	# Introduction
	In this blog post, we will explore how to solve the LeetCode problem "1808" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like [Chatmagic](https://www.chatmagic.app), we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

	# Problem Statement
	> You are given a positive integer primeFactors. You are asked to construct a positive integer n that satisfies the following conditions:  The number of prime factors of n (not necessarily distinct) is at most primeFactors. The number of nice divisors of n is maximized. Note that a divisor of n is nice if it is divisible by every prime factor of n. For example, if n = 12, then its prime factors are [2,2,3], then 6 and 12 are nice divisors, while 3 and 4 are not.  Return the number of nice divisors of n. Since that number can be too large, return it modulo 109 + 7. Note that a prime number is a natural number greater than 1 that is not a product of two smaller natural numbers. The prime factors of a number n is a list of prime numbers such that their product equals n.   Example 1: 
 Input: primeFactors = 5
 Output: 6
 Explanation: 200 is a valid value of n.
 It has 5 prime factors: [2,2,2,5,5], and it has 6 nice divisors: [10,20,40,50,100,200].
 There is not other value of n that has at most 5 prime factors and more nice divisors.
  Example 2: 
 Input: primeFactors = 8
 Output: 18
    Constraints:  1 <= primeFactors <= 109 

	# Explanation
	Here's the breakdown of the solution:

*   **Maximize exponents of 3:** To maximize the number of nice divisors, we need to maximize the product of exponents + 1 of the prime factors. Since we are given a limit on the *sum* of the exponents (primeFactors), we want to make the exponents as equal as possible. We achieve this by prioritizing the prime factor 3 since it provides the most "nice divisors" per prime factor used.
*   **Handle remainders with 2:** If the `primeFactors` is not divisible by 3, we take the remainder and distribute it to the exponents of 2 to fine-tune the number of nice divisors.
*   **Modular Exponentiation:** Efficiently calculate the powers needed, taking the modulo at each step to prevent overflow.

*   **Time Complexity:** O(log(primeFactors)) due to the power function.
*   **Space Complexity:** O(1)

	
	# Code
	```python
	def maxNiceDivisors(primeFactors: int) -> int:
    """
    Calculates the maximum number of nice divisors for a number n with at most primeFactors prime factors.

    Args:
        primeFactors: The maximum number of prime factors allowed.

    Returns:
        The maximum number of nice divisors modulo 10^9 + 7.
    """
    MOD = 10**9 + 7

    if primeFactors <= 3:
        return primeFactors

    num_threes = primeFactors // 3
    remainder = primeFactors % 3

    if remainder == 0:
        return pow(3, num_threes, MOD)
    elif remainder == 1:
        return (pow(3, num_threes - 1, MOD) * 4) % MOD
    else:  # remainder == 2
        return (pow(3, num_threes, MOD) * 2) % MOD
	```