Problem Description

You need to construct a positive integer n with specific properties to maximize a certain type of divisor count.

Given a positive integer primeFactors, construct an integer n such that:

1. Prime Factor Count Constraint: The total number of prime factors of n (counting multiplicities) must not exceed primeFactors. For example, if n = 12 = 2 × 2 × 3, it has 3 prime factors: [2, 2, 3].

2. Nice Divisor Definition: A divisor of n is called "nice" if it is divisible by every distinct prime that appears in n's prime factorization.

   • For example, if n = 12 = 2² × 3, the distinct primes are 2 and 3
   • The divisor 6 (= 2 × 3) is nice because it contains both primes
   • The divisor 12 (= 2² × 3) is also nice
   • But 3 and 4 are not nice divisors (3 doesn't contain prime 2, and 4 doesn't contain prime 3)

3. Objective: Maximize the number of nice divisors of n.

The problem asks you to return the maximum possible count of nice divisors. Since this number can be very large, return the result modulo 10⁹ + 7.

Key Insight: If n = p₁^k₁ × p₂^k₂ × ... × pₘ^kₘ where p₁, p₂, ..., pₘ are distinct primes, then:

• The sum k₁ + k₂ + ... + kₘ ≤ primeFactors
• The number of nice divisors equals k₁ × k₂ × ... × kₘ (since a nice divisor must have the form p₁^a₁ × p₂^a₂ × ... × pₘ^aₘ where 1 ≤ aᵢ ≤ kᵢ)

This transforms the problem into: how to split primeFactors into a sum of positive integers such that their product is maximized. The solution uses the mathematical principle that splitting into 3's (with special handling for remainders) gives the optimal product.

Intuition

The key realization is that we need to maximize the product of integers that sum to at most primeFactors. This is a classic mathematical optimization problem.

Let's think about how nice divisors are counted. If we have n = p₁^k₁ × p₂^k₂ × ... × pₘ^kₘ, then each nice divisor must include all distinct primes p₁, p₂, ..., pₘ. For each prime pᵢ, we can choose its exponent from 1 to kᵢ, giving us kᵢ choices. Therefore, the total number of nice divisors is k₁ × k₂ × ... × kₘ.

Now we face the question: given that k₁ + k₂ + ... + kₘ ≤ primeFactors, how do we maximize k₁ × k₂ × ... × kₘ?

This is equivalent to asking: how should we split a number into parts to maximize their product?

Let's explore with small examples:

• Splitting 6: 6 = 3 + 3 gives product 3 × 3 = 9, while 6 = 2 + 2 + 2 gives 2 × 2 × 2 = 8, and 6 = 1 + 5 gives only 1 × 5 = 5
• Splitting 8: 8 = 3 + 3 + 2 gives 3 × 3 × 2 = 18, while 8 = 2 + 2 + 2 + 2 gives 2 × 2 × 2 × 2 = 16

From mathematical analysis, we discover that:

1. We should avoid using 1 in our split (since multiplying by 1 doesn't increase the product)
2. We should prefer 3 over 2 when possible (since for the same sum, 3 + 3 = 6 gives product 9, while 2 + 2 + 2 = 6 gives product 8)
3. We shouldn't use numbers larger than 4 (since any number ≥ 5 can be split into 2 and 3 for a larger product)

The optimal strategy emerges:

• Use as many 3's as possible
• When primeFactors % 3 = 1, we have one leftover. Instead of using 3 + 1, we use 2 + 2 (since 2 × 2 = 4 > 3 × 1 = 3)
• When primeFactors % 3 = 2, we add one more 2 to our 3's

This explains why the solution computes 3^(primeFactors // 3) as the base and adjusts for remainders:

• Remainder 0: use all 3's
• Remainder 1: replace one 3 with two 2's (equivalent to multiplying by 4/3)
• Remainder 2: add one 2 to the product

Learn more about Recursion and Math patterns.

Solution Approach

The implementation follows the mathematical optimization strategy we discovered: splitting primeFactors into parts that maximize their product.

Step 1: Handle Base Cases For primeFactors < 4, we return primeFactors directly. This is because:

• If primeFactors = 1, we have n = p, giving 1 nice divisor
• If primeFactors = 2, we have n = p² or n = p₁ × p₂, both giving 2 nice divisors
• If primeFactors = 3, we have n = p³ or other combinations, with 3 being optimal

Step 2: Apply the Splitting Strategy For primeFactors ≥ 4, we examine primeFactors % 3:

Case 1: primeFactors % 3 == 0

• We can split primeFactors perfectly into groups of 3
• The maximum product is 3^(primeFactors // 3)
• We use pow(3, primeFactors // 3, mod) for efficient modular exponentiation

Case 2: primeFactors % 3 == 1

• We have primeFactors = 3k + 1 for some integer k
• Instead of using 3^k × 1, we use 3^(k-1) × 4
• This is because 3 + 1 = 4 can be split as 2 + 2, giving product 2 × 2 = 4
• The result is 4 × 3^(primeFactors // 3 - 1)

Case 3: primeFactors % 3 == 2

• We have primeFactors = 3k + 2 for some integer k
• We use 3^k × 2
• The extra 2 is added to the product of 3's
• The result is 2 × 3^(primeFactors // 3)

Step 3: Modular Arithmetic Since the result can be extremely large, we:

• Use Python's built-in pow(base, exp, mod) function for efficient modular exponentiation
• Apply % mod to the final result where mod = 10^9 + 7

The algorithm runs in O(log primeFactors) time due to the fast power operation, making it efficient even for large values of primeFactors.

Example Walkthrough

Let's walk through the solution with primeFactors = 10.

Step 1: Understanding what we're maximizing

• We need to construct an integer n using at most 10 prime factors (counting multiplicities)
• If n = p₁^k₁ × p₂^k₂ × ... × pₘ^kₘ, then k₁ + k₂ + ... + kₘ ≤ 10
• The number of nice divisors = k₁ × k₂ × ... × kₘ
• We need to maximize this product

Step 2: Finding the optimal split We need to split 10 into positive integers that maximize their product.

Let's consider different ways to split 10:

• 10 = 5 + 5: product = 5 × 5 = 25
• 10 = 4 + 3 + 3: product = 4 × 3 × 3 = 36
• 10 = 3 + 3 + 3 + 1: product = 3 × 3 × 3 × 1 = 27
• 10 = 3 + 3 + 2 + 2: product = 3 × 3 × 2 × 2 = 36
• 10 = 2 + 2 + 2 + 2 + 2: product = 2^5 = 32

Step 3: Applying the algorithm's strategy Since primeFactors = 10 and 10 % 3 = 1, we fall into Case 2:

• We would get 10 = 3 + 3 + 3 + 1 if we used all 3's
• But 3 × 1 = 3 is worse than 2 × 2 = 4
• So we replace one 3 and the 1 with two 2's
• This gives us: 10 = 3 + 3 + 2 + 2

Step 4: Calculate the result

• primeFactors // 3 = 3
• Since remainder is 1, we use: 4 × 3^(3-1) = 4 × 3^2 = 4 × 9 = 36

Step 5: Construct the actual n One possible n would be:

• n = 2³ × 3³ × 5² × 7² (using exponents 3, 3, 2, 2 which sum to 10)
• Nice divisors must include all four primes: 2, 3, 5, and 7
• For each prime, we can choose exponents: 1-3 for 2, 1-3 for 3, 1-2 for 5, 1-2 for 7
• Total nice divisors = 3 × 3 × 2 × 2 = 36

This confirms our algorithm correctly identifies that the maximum number of nice divisors is 36 when primeFactors = 10.

Solution Implementation

Time and Space Complexity

The time complexity of this algorithm is O(log n), where n is the value of primeFactors. This is because the dominant operation in the code is the pow(3, primeFactors // 3, mod) function call, which uses modular exponentiation. The built-in pow function with three arguments performs fast exponentiation using the square-and-multiply algorithm, which takes O(log k) time where k is the exponent. Since the exponent is at most primeFactors // 3, and in the worst case could be approximately primeFactors / 3, the time complexity is O(log(primeFactors)) or O(log n).

The space complexity is O(1) as the algorithm only uses a constant amount of extra space. The variables used are mod and the temporary values during computation. The modular exponentiation operation pow(3, primeFactors // 3, mod) is implemented iteratively in Python's built-in function and doesn't require additional space proportional to the input size.

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Integer Overflow in Multiplication

One of the most common mistakes is forgetting to apply modulo operations during intermediate multiplication steps, especially in the cases where remainder is 1 or 2.

Incorrect Implementation:

# This can cause overflow for large quotients
if remainder == 1:
    return 4 * pow(3, quotient - 1, MOD) % MOD  # Overflow risk!

Correct Implementation:

# Apply modulo to prevent overflow
if remainder == 1:
    return (4 * pow(3, quotient - 1, MOD)) % MOD

2. Misunderstanding the Remainder == 1 Case

Many solutions incorrectly handle when primeFactors % 3 == 1. The intuition that "just add 1 to the product" is wrong because 3^k × 1 = 3^k, which doesn't utilize the extra prime factor.

Incorrect Approach:

if remainder == 1:
    return pow(3, quotient, MOD)  # Wrong! This ignores the extra 1

Correct Approach:

if remainder == 1:
    # Take one 3 away and combine with the 1 to make 4 (as 2×2)
    return (4 * pow(3, quotient - 1, MOD)) % MOD

3. Using Regular Power Instead of Modular Exponentiation

For large values of primeFactors, computing 3^quotient directly will cause overflow or timeout.

Incorrect:

result = (3 ** quotient) % MOD  # Too slow and can overflow!

Correct:

result = pow(3, quotient, MOD)  # Uses fast modular exponentiation

4. Off-by-One Error in Base Cases

Some implementations incorrectly handle small values of primeFactors, particularly forgetting that primeFactors = 4 should be split as 2 + 2 (product = 4), not returned as 4 directly.

Incorrect Base Case:

if primeFactors <= 4:
    return primeFactors  # Wrong for primeFactors = 4

Correct Base Case:

if primeFactors < 4:
    return primeFactors
# primeFactors = 4 should go through the main logic

5. Not Considering Edge Case of primeFactors = 0

While not typically part of the problem constraints, handling primeFactors = 0 incorrectly can cause issues.

Better Implementation with Guard:

def maxNiceDivisors(self, primeFactors: int) -> int:
    if primeFactors == 0:
        return 0  # or 1, depending on problem definition
  
    MOD = 10**9 + 7
    if primeFactors < 4:
        return primeFactors
    # ... rest of the logic