Maximize Number of Nice Divisors

Hard

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Greedy AlgorithmsBit Manipulation

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

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum possible value of primeFactors?
Can primeFactors be zero or negative?
If primeFactors is a small number like 1 or 2, can you provide an example of what a 'nice divisor' would be for n in those cases to ensure I understand the definition correctly?
Are we only concerned with maximizing the *number* of nice divisors, or are there any additional criteria (e.g., maximizing the value of the nice divisors)?
Can you elaborate on the relationship between primeFactors and the prime factors of n itself? Specifically, does primeFactors represent the *distinct* number of prime factors, or the *total* count of prime factors when considering multiplicity?

Brute Force Solution

Approach

The problem asks us to find the largest product we can make by breaking down a number. The brute force method is simply trying every possible way to break down the number and calculating the resulting product each time.

Here's how the algorithm would work step-by-step:

Start by considering all possible ways to split the original number into smaller whole numbers.
For each possible split, multiply all the smaller numbers together to get a product.
Compare all of these products, and keep track of the biggest one you've found so far.
Once you've tried every possible way to split the original number, the largest product you kept track of is your answer.

Code Implementation

def maximize_number_nice_divisors_brute_force(prime_factors):
    max_product = 0

    def calculate_product(splits):
        product = 1
        for split_value in splits:
            product *= split_value
        return product

    def find_all_splits(remaining_prime_factors, current_splits):
        nonlocal max_product
        
        # If no prime factors remain, calculate and compare the product
        if remaining_prime_factors == 0:
            product = calculate_product(current_splits)
            max_product = max(max_product, product)
            return
        
        # Ensure we don't go negative
        for i in range(1, min(4, remaining_prime_factors + 1)):
            find_all_splits(remaining_prime_factors - i, current_splits + [i])

    # Start the recursive process
    find_all_splits(prime_factors, [])
    return max_product

Big(O) Analysis

Time Complexity

O(n!) – The provided brute force approach involves considering all possible ways to split the original number 'n' into smaller whole numbers. This essentially translates to generating all possible compositions of 'n'. The number of compositions of an integer 'n' grows factorially with 'n'. Therefore, the algorithm must iterate through an exponentially increasing number of partitions or splits. Each split also requires multiplication of the split values, which takes O(n) in the worst case (when there are n numbers to multiply). Thus the overall time complexity grows approximately with n!, making it O(n!).

Space Complexity

O(N) – The brute force method, as described, explores all possible ways to split the original number N. In the worst case, to generate each possible split and calculate the product, we might need to temporarily store the split numbers in a data structure. The size of this temporary storage could grow linearly with N, as in a scenario where we are splitting N into N ones. Therefore, the space complexity is O(N), driven by the temporary storage required to generate and process each split. Although there is no explicit mention of a data structure, this implicit storage to hold split numbers while calculating products is necessary in the brute force steps described.

Optimal Solution

Approach

The key to maximizing the number of nice divisors is understanding that we want to break down the initial number into as many 3s as possible. Using some math properties allows us to determine the optimal combination of 3s and sometimes a 2 to achieve the highest product.

Here's how the algorithm would work step-by-step:

First, handle the simple cases where the input is 1, 2, or 3, since these have direct answers.
For larger inputs, repeatedly divide the number by 3 as many times as possible, keeping track of how many 3s we have.
If, after dividing by 3s, you are left with 1, then reduce the number of 3s by 1 and multiply by 4, because 3 * 1 is worse than 2 * 2.
If you are left with 2, then multiply your 3s by the single 2 to complete the calculation.
Finally, because the answer can be very large, calculate the result while ensuring that all calculations are performed modulo a certain number (10^9 + 7).

Code Implementation

def maximize_number_of_nice_divisors(prime_factorization_exponent):
    modulo = 10**9 + 7

    if prime_factorization_exponent <= 3:
        return prime_factorization_exponent

    number_of_threes = prime_factorization_exponent // 3
    remainder = prime_factorization_exponent % 3

    # If remainder is 0, we only need to consider powers of 3
    if remainder == 0:
        result = pow(3, number_of_threes, modulo)

    # If remainder is 1, replace one 3 with two 2s (3*1 < 2*2)
    elif remainder == 1:
        number_of_threes -= 1
        result = (pow(3, number_of_threes, modulo) * 4) % modulo

    # If remainder is 2, include it as is (3*2 > other options)
    else:
        result = (pow(3, number_of_threes, modulo) * 2) % modulo

    return result

Big(O) Analysis

Time Complexity

O(log n) – The dominant operation is repeatedly dividing the input number n by 3. The number of iterations this division occurs is proportional to how many times we can divide n by 3 until we reach a base case (1, 2, or no remainder). This effectively represents a logarithmic operation with base 3, meaning the loop runs approximately log₃(n) times. All other operations (multiplication, conditional checks) take constant time, and are performed either once, or a limited number of times. Thus, the overall time complexity is O(log n).

Space Complexity

O(1) – The algorithm uses a fixed number of variables to store the count of 3s and potentially a remainder (1 or 2). No auxiliary data structures like arrays or hash maps are created. The space used does not depend on the input value N, so the space complexity is constant.

Edge Cases

Case	How to Handle
primeFactors is 1	Return 1 because the number n is a single prime, hence one nice divisor which is the number itself.
primeFactors is 2	Return primeFactors because number n is product of two primes.
primeFactors is 3	Return 4 since n is p1p2p3 and nice divisors are p1p2p3 and products of two elements
primeFactors is a large number, causing integer overflow during exponentiation.	Use modular exponentiation to prevent integer overflow by applying modulo operation after each multiplication step.
primeFactors is 0	Return 0 since the number is 1, and it has no nice divisors.
primeFactors is a large number where all divisions result in numbers near 0.	Optimize division to calculate powers efficiently using binary exponentiation while managing modular arithmetic.
Input can result in a result that will overflow even the long data type	Modular arithmetic should be used during all intermediate calculations to avoid overflow.
Optimal split between 2s and 3s may not be obvious	Try splitting into as many 3s as possible then distributing remaining factors as 2s and check the product.

primeFactors is 1

How to Handle:

Return 1 because the number n is a single prime, hence one nice divisor which is the number itself.

primeFactors is 2

How to Handle:

Return primeFactors because number n is product of two primes.

primeFactors is 3

How to Handle:

Return 4 since n is p1*p2*p3 and nice divisors are p1*p2*p3 and products of two elements

primeFactors is a large number, causing integer overflow during exponentiation.

How to Handle:

Use modular exponentiation to prevent integer overflow by applying modulo operation after each multiplication step.

primeFactors is 0

How to Handle:

Return 0 since the number is 1, and it has no nice divisors.

primeFactors is a large number where all divisions result in numbers near 0.

How to Handle:

Optimize division to calculate powers efficiently using binary exponentiation while managing modular arithmetic.

Input can result in a result that will overflow even the long data type

How to Handle:

Modular arithmetic should be used during all intermediate calculations to avoid overflow.

Optimal split between 2s and 3s may not be obvious

How to Handle:

Try splitting into as many 3s as possible then distributing remaining factors as 2s and check the product.