LeetCode #1808 — HARD

Maximize Number of Nice Divisors

Break down a hard problem into reliable checkpoints, edge-case handling, and complexity trade-offs.

The Problem

Problem Statement

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

Step 01

Brute Force Baseline

Problem summary: 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.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Math

Example 1

Example 2

Core Insight

What unlocks the optimal approach

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.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Largest constraint values

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
class Solution {
    private final int mod = (int) 1e9 + 7;

    public int maxNiceDivisors(int primeFactors) {
        if (primeFactors < 4) {
            return primeFactors;
        }
        if (primeFactors % 3 == 0) {
            return qpow(3, primeFactors / 3);
        }
        if (primeFactors % 3 == 1) {
            return (int) (4L * qpow(3, primeFactors / 3 - 1) % mod);
        }
        return 2 * qpow(3, primeFactors / 3) % mod;
    }

    private int qpow(long a, long n) {
        long ans = 1;
        for (; n > 0; n >>= 1) {
            if ((n & 1) == 1) {
                ans = ans * a % mod;
            }
            a = a * a % mod;
        }
        return (int) ans;
    }
}

// Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
func maxNiceDivisors(primeFactors int) int {
	if primeFactors < 4 {
		return primeFactors
	}
	const mod = 1e9 + 7
	qpow := func(a, n int) int {
		ans := 1
		for ; n > 0; n >>= 1 {
			if n&1 == 1 {
				ans = ans * a % mod
			}
			a = a * a % mod
		}
		return ans
	}
	if primeFactors%3 == 0 {
		return qpow(3, primeFactors/3)
	}
	if primeFactors%3 == 1 {
		return qpow(3, primeFactors/3-1) * 4 % mod
	}
	return qpow(3, primeFactors/3) * 2 % mod
}

# Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
class Solution:
    def maxNiceDivisors(self, primeFactors: int) -> int:
        mod = 10**9 + 7
        if primeFactors < 4:
            return primeFactors
        if primeFactors % 3 == 0:
            return pow(3, primeFactors // 3, mod) % mod
        if primeFactors % 3 == 1:
            return 4 * pow(3, primeFactors // 3 - 1, mod) % mod
        return 2 * pow(3, primeFactors // 3, mod) % mod

// Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
/**
 * [1808] Maximize Number of Nice Divisors
 *
 * 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 10^9 + 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 <= 10^9
 *
 */
pub struct Solution {}

// problem: https://leetcode.com/problems/maximize-number-of-nice-divisors/
// discuss: https://leetcode.com/problems/maximize-number-of-nice-divisors/discuss/?currentPage=1&orderBy=most_votes&query=

// submission codes start here
const MOD: i64 = 1_000_000_007;

impl Solution {
    // Credit: https://leetcode.com/problems/maximize-number-of-nice-divisors/solutions/3220691/just-a-runnable-solution/
    pub fn max_nice_divisors(prime_factors: i32) -> i32 {
        let prime_factors = prime_factors as i64;

        let st = [0_i64, 1, 2, 3, 4, 6];
        if prime_factors < 6 {
            return st[prime_factors as usize] as _;
        }

        let mut result = st[3 + prime_factors as usize % 3];
        let mut base = 3;
        let mut exp = prime_factors / 3 - 1;
        while exp > 0 {
            if exp & 1 == 1 {
                result = result * base % MOD;
            }
            base = base * base % MOD;
            exp >>= 1;
        }

        result as _
    }
}

// submission codes end

#[cfg(test)]
mod tests {
    use super::*;

    #[test]
    fn test_1808_example_1() {
        let prime_factors = 5;

        let result = 6;

        assert_eq!(Solution::max_nice_divisors(prime_factors), result);
    }

    #[test]
    fn test_1808_example_2() {
        let prime_factors = 8;

        let result = 18;

        assert_eq!(Solution::max_nice_divisors(prime_factors), result);
    }
}

// Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
// Auto-generated TypeScript example from java.
function exampleSolution(): void {
}
// Reference (java):
// // Accepted solution for LeetCode #1808: Maximize Number of Nice Divisors
// class Solution {
//     private final int mod = (int) 1e9 + 7;
// 
//     public int maxNiceDivisors(int primeFactors) {
//         if (primeFactors < 4) {
//             return primeFactors;
//         }
//         if (primeFactors % 3 == 0) {
//             return qpow(3, primeFactors / 3);
//         }
//         if (primeFactors % 3 == 1) {
//             return (int) (4L * qpow(3, primeFactors / 3 - 1) % mod);
//         }
//         return 2 * qpow(3, primeFactors / 3) % mod;
//     }
// 
//     private int qpow(long a, long n) {
//         long ans = 1;
//         for (; n > 0; n >>= 1) {
//             if ((n & 1) == 1) {
//                 ans = ans * a % mod;
//             }
//             a = a * a % mod;
//         }
//         return (int) ans;
//     }
// }

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(log n)

Space

O(1)

Approach Breakdown

ITERATIVE

O(n) time

O(1) space

Simulate the process step by step — multiply n times, check each number up to n, or iterate through all possibilities. Each step is O(1), but doing it n times gives O(n). No extra space needed since we just track running state.

MATH INSIGHT

O(log n) time

O(1) space

Math problems often have a closed-form or O(log n) solution hidden behind an O(n) simulation. Modular arithmetic, fast exponentiation (repeated squaring), GCD (Euclidean algorithm), and number theory properties can dramatically reduce complexity.

Shortcut: Look for mathematical properties that eliminate iteration. Repeated squaring → O(log n). Modular arithmetic avoids overflow.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Overflow in intermediate arithmetic

Wrong move: Temporary multiplications exceed integer bounds.

Usually fails on: Large inputs wrap around unexpectedly.

Fix: Use wider types, modular arithmetic, or rearranged operations.