How do you solve Maximize Number of Nice Divisors on LeetCode?

Maximize Number of Nice Divisors (LeetCode #1808) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(1) time complexity and O(1) space complexity. Key insight: Using only 2s and 3s maximizes the product of numbers that sum to primeFactors.

What is the brute force solution for Maximize Number of Nice Divisors?

The brute force approach for Maximize Number of Nice Divisors is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves generating all possible combinations of prime factors up to the given limit and calculating the number of nice divisors for each combination. This is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(1).

What algorithm or data structure is used to solve Maximize Number of Nice Divisors?

Maximize Number of Nice Divisors uses the following concepts: Math, Recursion, Number Theory. The recommended patterns to study are: Greedy Algorithms, Mathematical Optimization.

What are the interview tips for Maximize Number of Nice Divisors?

Always think about how to break down larger problems into smaller parts. Be ready to explain why certain numbers (like 2 and 3) are optimal in this context. Practice modular arithmetic to handle large numbers effectively.

What are common mistakes when solving Maximize Number of Nice Divisors?

Not considering the case when primeFactors is 1. Failing to optimize the use of 2s and 3s, leading to suboptimal products.

#1808

Maximize Number of Nice Divisors

Hard

MathRecursionNumber TheoryGreedy AlgorithmsMathematical Optimization

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(1)
Space	O(1)	O(1)

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Intuition

Time O(1)Space O(1)

The optimal solution leverages the fact that to maximize the product of numbers that sum to primeFactors, we should use 2s and 3s. This is based on the mathematical property that breaking larger numbers into smaller parts increases the product.

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Algorithm

4 steps

1Step 1: If primeFactors is 1, return 1 since the only number is 1.
2Step 2: Calculate how many 3s can fit into primeFactors.
3Step 3: If there's a remainder of 1 after using 3s, replace one 3 with two 2s to maximize the product.
4Step 4: Calculate the product of the 2s and 3s, and return the result modulo 10^9 + 7.

solution.py13 lines

1# Full working Python code
2MOD = 10**9 + 7
3
4def maxNiceDivisors(primeFactors):
5    if primeFactors == 1:
6        return 1
7    count3 = primeFactors // 3
8    remainder = primeFactors % 3
9    if remainder == 1:
10        count3 -= 1
11        remainder += 3
12    count2 = remainder // 2
13    return pow(3, count3, MOD) * pow(2, count2, MOD) % MOD

ℹ

Complexity note: The optimal solution runs in constant time O(1) since it only involves a few arithmetic operations and modular exponentiation, regardless of the size of primeFactors.

1Using only 2s and 3s maximizes the product of numbers that sum to primeFactors.
2If you have a remainder of 1 when dividing by 3, adjust by replacing one 3 with two 2s.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.