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Maximize Number of Nice Divisors

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

Solution Explanation for Maximize Number of Nice Divisors

This problem asks to find a positive integer n with at most primeFactors prime factors (not necessarily distinct) such that the number of its "nice divisors" is maximized. A nice divisor is a divisor of n that's divisible by every prime factor of n.

The core idea lies in recognizing the pattern of how to maximize the number of nice divisors. Let's analyze:

1. Problem Transformation:

If we represent n as its prime factorization: n = p1^a1 * p2^a2 * ... * pk^ak, where pi are distinct primes and ai are their exponents, then a nice divisor must contain at least one of each pi. The number of nice divisors is then determined by the exponents a1, a2, ..., ak. The total number of prime factors is a1 + a2 + ... + ak <= primeFactors. The number of nice divisors is a1 * a2 * ... * ak.

Therefore, the problem transforms into finding non-negative integers a1, a2, ..., ak that satisfy a1 + a2 + ... + ak <= primeFactors and maximize the product a1 * a2 * ... * ak.

2. Optimizing the Product:

Intuitively, to maximize the product, we want to have as many factors as possible, with values as close to each other as possible. It turns out that using 3's is the best strategy. Consider these scenarios:

  • Using 3s: If you can use a multiple of 3 (say, 3k), it's better to break it up into 3s because 3 * 3 > 4 (and 3 * 3 * 3 > 2 * 2 * 2 * 2).
  • Dealing with remainders: If primeFactors isn't divisible by 3:
    • If it has a remainder of 1 when divided by 3, it is more efficient to have one 4 and remaining 3s.
    • If it has a remainder of 2 when divided by 3, one 2 and the rest 3s are optimal.

3. Code Implementation:

The code across all languages implements this optimization:

  1. Base Cases: If primeFactors is less than 4, the maximum number of nice divisors is simply primeFactors.
  2. Divisible by 3: If primeFactors is divisible by 3, the maximum number of nice divisors is 3 raised to the power of (primeFactors/3). This is calculated efficiently using fast exponentiation (qpow function) to handle large exponents and modulo operations to avoid integer overflow.
  3. Remainder 1: If primeFactors has a remainder of 1 when divided by 3, we use one 4 and the rest as 3s.
  4. Remainder 2: If primeFactors has a remainder of 2 when divided by 3, we use one 2 and the rest as 3s.

Time and Space Complexity:

  • Time Complexity: O(log primeFactors) due to the fast exponentiation (qpow) function. The rest of the operations are constant time.
  • Space Complexity: O(1). The code uses a constant amount of extra space.

Example Walkthrough (primeFactors = 8):

  1. 8 is not less than 4.
  2. 8 % 3 == 2.
  3. The algorithm calculates 2 * 3^(8/3) = 2 * 3^2 = 18. This is the maximum number of nice divisors.

The provided code in Python, Java, C++, Go, and Javascript all efficiently implement this logic, utilizing fast exponentiation to manage potential overflow issues with large inputs.