Maximize Number of Nice Divisors - Practice Coding Problems

Maximize Number of Nice Divisors - Problem

Maximize Number of Nice Divisors

You're given a positive integer primeFactors representing the maximum number of prime factors you can use to construct a number n. Your goal is to maximize the number of "nice divisors" of n.

A nice divisor is a divisor that is divisible by every prime factor of n. For example, if n = 12 with prime factors [2,2,3], then 6 and 12 are nice divisors (both are divisible by 2 and 3), while 3 and 4 are not.

Key Insight: The number of nice divisors equals the product of exponents of distinct prime factors. To maximize this product with a fixed sum (primeFactors), we need to optimally distribute the exponents.

Return: The maximum number of nice divisors modulo 10⁹ + 7.

Input & Output

example_1.py — Basic Case

$ Input: primeFactors = 5

› Output: 6

💡 Note: One optimal way is to construct n = 2^3 × 3^2 (using 5 prime factors). The nice divisors are those divisible by both 2 and 3: [6, 12, 18, 36, 72, 108]. But the actual count comes from the exponents: 3 × 2 = 6. The optimal distribution is [3, 2] giving product 6.

example_2.py — Divisible by 3

$ Input: primeFactors = 9

› Output: 27

💡 Note: Since 9 is divisible by 3, we can use three groups of 3: [3, 3, 3]. This gives us the maximum product of 3 × 3 × 3 = 27.

example_3.py — Remainder 1

$ Input: primeFactors = 10

› Output: 36

💡 Note: When primeFactors % 3 == 1, we use one less group of 3 and instead use two groups of 2. So instead of [3, 3, 3, 1] (product = 27), we use [3, 3, 2, 2] (product = 36), which is better.

Constraints

1 ≤ primeFactors ≤ 10⁹
Return result modulo 10⁹ + 7
primeFactors represents the maximum total count of prime factors (with repetition)

Visualization

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Understanding the Visualization

Equal Distribution

Start by trying to distribute candies as evenly as possible

Magic Number 3

Discover that groups of 3 candies give better results than other groupings

Handle Remainders

When you can't make perfect groups of 3, optimize the remaining candies

Calculate Result

Use efficient multiplication to get the final happiness score

Key Takeaway

🎯 Key Insight: Use the mathematical principle that to maximize a product with a fixed sum, make all factors as close to e ≈ 2.718 as possible. Since 3 is the closest integer to e, we use as many 3s as possible.

Asked in

G Google 23 a Amazon 18 ⊞ Microsoft 15 f Meta 12

The optimal solution uses mathematical insights from AM-GM inequality: to maximize a product with fixed sum, make factors as equal as possible. Since 3 is closest to e ≈ 2.718, we prefer 3s over other factors. Use fast exponentiation to compute 3^k efficiently in O(log n) time.

Common Approaches

Approach	Time	Space	Notes
✓ Brute Force (Recursive Enumeration)	O(2^n)	O(n)	Try all possible ways to partition primeFactors and find maximum product
Mathematical Analysis + Greedy	O(log n)	O(1)	First analyze the optimal distribution pattern, then apply greedy strategy

Brute Force (Recursive Enumeration) — Algorithm Steps

Generate all partitions of primeFactors
For each partition, calculate the product of all parts
Keep track of the maximum product found
Return the maximum modulo 10^9 + 7

Visualization

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Step-by-Step Walkthrough

Root Node

Start with the full number to partition

Branch Creation

For each possible first part, create a branch

Recursive Descent

Recursively partition the remainder

Product Calculation

Calculate product for each complete partition

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>

#define MOD 1000000007

int maxProduct = 0;

void findMaxProduct(int remaining, int currentProduct, int lastUsed) {
    if (remaining == 0) {
        if (currentProduct > maxProduct) {
            maxProduct = currentProduct;
        }
        return;
    }
    
    for (int i = lastUsed; i <= remaining; i++) {
        long long newProduct = ((long long)currentProduct * i) % MOD;
        findMaxProduct(remaining - i, (int)newProduct, i);
    }
}

int maxNiceDivisors(int primeFactors) {
    maxProduct = 0;
    findMaxProduct(primeFactors, 1, 1);
    return maxProduct;
}

int main() {
    int primeFactors;
    scanf("%d", &primeFactors);
    printf("%d\n", maxNiceDivisors(primeFactors));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(2^n)

Number of partitions grows exponentially with n

⚠ Quadratic Growth

Space Complexity

O(n)

Recursion stack depth and storage for current partition

⚡ Linearithmic Space

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Medium Frequency

~25 min Avg. Time

892 Likes

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Smart Actions

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Input & Output

Constraints

Visualization

Related Problems

Common Approaches

Brute Force (Recursive Enumeration) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

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