You are given a positive integer primeFactors. You are asked to construct a positive integer n that satisfies the following conditions:
n (not necessarily distinct) is at most primeFactors.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 <= 109In #1808 Maximize Number of Nice Divisors, the goal is to construct a number using a limited count of prime factors such that the number of nice divisors (divisors divisible by every prime factor) is maximized. Instead of brute‑forcing factorizations, the key observation comes from mathematical optimization: the number of nice divisors depends on how prime factors are grouped into exponents.
To maximize the product formed by these exponent groups, the problem reduces to a classic integer break strategy. Splitting the total number of prime factors into smaller groups—especially around 3—maximizes the resulting multiplicative value. This allows the solution to be computed using fast modular exponentiation rather than recursion or enumeration.
Because the result can be extremely large, calculations are performed under a modulus using modular power. This mathematical reduction turns a potentially exponential search into a very efficient computation.
Time Complexity: O(log n) due to fast exponentiation.
Space Complexity: O(1).
| Approach | Time Complexity | Space Complexity |
|---|---|---|
| Mathematical optimization with modular exponentiation | O(log n) | O(1) |
NeetCode
Use these hints if you're stuck. Try solving on your own first.
The number of nice divisors is equal to the product of the count of each prime factor. Then the problem is reduced to: given n, find a sequence of numbers whose sum equals n and whose product is maximized.
This sequence can have no numbers that are larger than 4. Proof: if it contains a number x that is larger than 4, then you can replace x with floor(x/2) and ceil(x/2), and floor(x/2) * ceil(x/2) > x. You can also replace 4s with two 2s. Hence, there will always be optimal solutions with only 2s and 3s.
If there are three 2s, you can replace them with two 3s to get a better product. Hence, you'll never have more than two 2s.
Keep adding 3s as long as n ≥ 5.
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Mathematically, breaking a number into factors of 3 tends to maximize the product compared to other splits. This principle is commonly used in integer break problems and helps maximize the number of nice divisors generated from the available prime factors.
Yes, variations of this problem can appear in FAANG-style interviews because it tests mathematical insight, number theory reasoning, and optimization techniques. Candidates are expected to recognize the connection to integer break and implement efficient modular exponentiation.
This problem primarily relies on mathematical reasoning rather than complex data structures. The implementation typically uses integer arithmetic and modular exponentiation, so no specialized data structure is required.
The optimal approach uses mathematical optimization similar to the integer break problem. By splitting the available prime factors into groups that maximize the product, the number of nice divisors can be maximized. Fast modular exponentiation is then used to compute the result efficiently.