LeetCode 1808 - Maximize Number of Nice Divisors

The problem gives us an integer primeFactors, representing the maximum total number of prime factors we are allowed to use when constructing some positive integer n. The goal is not to maximize n itself. Instead, we want to maximize the number of "nice divisors" of n.

leetcodehardmathrecursionnumber-theory

LeetCode Problem 1808

Difficulty: 🔴 Hard
Topics: Math, Recursion, Number Theory

Solution

Problem Understanding

The problem gives us an integer primeFactors, representing the maximum total number of prime factors we are allowed to use when constructing some positive integer n.

The goal is not to maximize n itself. Instead, we want to maximize the number of "nice divisors" of n.

A divisor of n is considered nice if it is divisible by every distinct prime factor of n.

Suppose:

  • n = 2^a * 3^b * 5^c

Then any nice divisor must contain at least one copy of every distinct prime factor:

  • at least one 2
  • at least one 3
  • at least one 5

The number of such divisors becomes:

$$a \times b \times c$$

because:

  • the divisor may choose exponent 1..a for prime 2
  • exponent 1..b for prime 3
  • exponent 1..c for prime 5

So the entire problem transforms into:

Split primeFactors into positive integers whose product is maximized.

This is the same mathematical optimization problem as the classic "integer break" problem.

The constraints are extremely large:

  • 1 <= primeFactors <= 10^9

This immediately tells us:

  • brute force enumeration is impossible
  • dynamic programming over all values is also impossible
  • we need a mathematical solution with logarithmic complexity

Several edge cases are important:

  • primeFactors = 1
  • primeFactors = 2
  • primeFactors = 3

These small values do not follow the normal "split into 3s" rule and must be handled carefully.

Another important issue is overflow. The answer can become astronomically large, so all calculations must be performed modulo:

$$10^9 + 7$$

Efficient modular exponentiation is therefore necessary.

Approaches

Brute Force Approach

A brute force solution would attempt to enumerate all possible ways to distribute the available prime factors among different prime powers.

For example, if primeFactors = 8, we could try partitions like:

  • 8
  • 7 + 1
  • 6 + 2
  • 5 + 3
  • 4 + 4
  • 3 + 3 + 2
  • etc.

For each partition:

  • interpret each part as the exponent of a distinct prime
  • compute the product of the parts
  • keep the maximum

This works because the number of nice divisors equals the product of the exponents.

However, the number of integer partitions grows exponentially. With primeFactors up to 10^9, brute force is completely infeasible.

Key Insight

The important mathematical observation is:

To maximize the product of integers whose sum is fixed, we should use as many 3s as possible.

This is a well known optimization result.

Why 3?

  • Splitting a number into larger chunks reduces the product.
  • Splitting into too many 2s is less efficient than 3s.
  • Any number >= 5 can be improved by extracting a 3.

For example:

$$5 = 2 + 3,\quad 2 \times 3 = 6 > 5$$

$$6 = 3 + 3,\quad 3 \times 3 = 9 > 6$$

The only special case is when the remainder becomes 1.

For example:

$$10 = 3 + 3 + 3 + 1$$

This is bad because:

$$3 \times 3 \times 3 \times 1 = 27$$

Instead:

$$10 = 3 + 3 + 4$$

which gives:

$$3 \times 3 \times 4 = 36$$

So the optimal strategy becomes:

  • Use as many 3s as possible.
  • If remainder is 1, convert one 3 + 1 into 2 + 2.
  • If remainder is 2, keep it.

Approach Comparison

Approach Time Complexity Space Complexity Notes
Brute Force Exponential Exponential Enumerates integer partitions
Optimal O(log n) O(1) Uses mathematical decomposition and fast exponentiation

Algorithm Walkthrough

Step 1: Handle Small Values

If primeFactors <= 3, return primeFactors.

Why?

Because splitting them further does not improve the product.

Examples:

  • 2 -> 2
  • 3 -> 3

Step 2: Divide by 3

Compute:

$$q = primeFactors // 3$$

$$r = primeFactors % 3$$

The quotient tells us how many 3s we can use.

The remainder determines the adjustment strategy.

Step 3: Handle Remainder 0

If:

$$r = 0$$

then the optimal decomposition is entirely 3s.

Result:

$$3^q$$

Step 4: Handle Remainder 1

If:

$$r = 1$$

we should avoid ending with +1.

Instead of:

$$3^{q} \times 1$$

we transform:

$$3 + 1 \rightarrow 2 + 2$$

So:

$$3^{q-1} \times 4$$

Step 5: Handle Remainder 2

If:

$$r = 2$$

simply append a 2.

Result:

$$3^q \times 2$$

Step 6: Use Modular Exponentiation

Since powers become extremely large, compute them modulo:

$$10^9 + 7$$

Python provides:

pow(base, exponent, mod)

which performs fast exponentiation in logarithmic time.

In Go, we implement binary exponentiation manually.

Why it works

The correctness comes from the mathematical property that breaking a number into mostly 3s maximizes the product for a fixed sum.

Any integer >= 5 benefits from splitting off a 3:

$$x < 3(x-3)$$

for all x >= 5.

Thus repeatedly extracting 3s always improves the product until only 2, 3, or 4 remain.

The only problematic remainder is 1, because multiplying by 1 contributes nothing. Replacing 3 + 1 with 2 + 2 increases the product:

$$3 \times 1 < 2 \times 2$$

Therefore the algorithm always produces the optimal decomposition.

Python Solution

class Solution:
    def maxNiceDivisors(self, primeFactors: int) -> int:
        MOD = 10**9 + 7

        if primeFactors <= 3:
            return primeFactors

        quotient = primeFactors // 3
        remainder = primeFactors % 3

        if remainder == 0:
            return pow(3, quotient, MOD)

        if remainder == 1:
            return (pow(3, quotient - 1, MOD) * 4) % MOD

        return (pow(3, quotient, MOD) * 2) % MOD

The implementation follows the mathematical derivation directly.

The first section handles the small edge cases where splitting is not beneficial.

Next, we divide the value by 3 to determine how many groups of 3 should appear in the optimal partition.

The remainder determines which formula to use:

  • remainder 0 means all 3s
  • remainder 1 means replace one 3 + 1 with 2 + 2
  • remainder 2 means append a 2

The built in pow(base, exponent, mod) function performs efficient modular exponentiation in logarithmic time.

Go Solution

func maxNiceDivisors(primeFactors int) int {
	const MOD int64 = 1_000_000_007

	if primeFactors <= 3 {
		return primeFactors
	}

	quotient := primeFactors / 3
	remainder := primeFactors % 3

	if remainder == 0 {
		return int(modPow(3, int64(quotient), MOD))
	}

	if remainder == 1 {
		return int((modPow(3, int64(quotient-1), MOD) * 4) % MOD)
	}

	return int((modPow(3, int64(quotient), MOD) * 2) % MOD)
}

func modPow(base, exponent, mod int64) int64 {
	result := int64(1)

	base %= mod

	for exponent > 0 {
		if exponent&1 == 1 {
			result = (result * base) % mod
		}

		base = (base * base) % mod
		exponent >>= 1
	}

	return result
}

The Go implementation mirrors the Python logic closely.

Since Go does not provide a built in modular exponentiation function for integers, we implement binary exponentiation manually in modPow.

All arithmetic is performed using int64 to avoid overflow during multiplication before taking modulo.

The final answer is converted back to int because LeetCode expects that return type.

Worked Examples

Example 1

Input:

primeFactors = 5

Compute:

Variable Value
quotient 1
remainder 2

Since remainder is 2:

$$3^1 \times 2$$

Result:

$$3 \times 2 = 6$$

Final answer:

6

Optimal partition:

$$5 = 3 + 2$$

Example 2

Input:

primeFactors = 8

Compute:

Variable Value
quotient 2
remainder 2

Since remainder is 2:

$$3^2 \times 2$$

Result:

$$9 \times 2 = 18$$

Final answer:

18

Optimal partition:

$$8 = 3 + 3 + 2$$

Product:

$$3 \times 3 \times 2 = 18$$

Complexity Analysis

Measure Complexity Explanation
Time O(log n) Modular exponentiation uses binary exponentiation
Space O(1) Only constant extra variables are used

The dominant operation is exponentiation.

Binary exponentiation repeatedly squares the base while halving the exponent, so the number of iterations is proportional to:

$$\log(primeFactors)$$

No auxiliary data structures are required.

Test Cases

sol = Solution()

assert sol.maxNiceDivisors(1) == 1  # minimum input
assert sol.maxNiceDivisors(2) == 2  # small edge case
assert sol.maxNiceDivisors(3) == 3  # small edge case
assert sol.maxNiceDivisors(4) == 4  # 2 + 2
assert sol.maxNiceDivisors(5) == 6  # example case
assert sol.maxNiceDivisors(6) == 9  # 3 + 3
assert sol.maxNiceDivisors(7) == 12  # 3 + 2 + 2
assert sol.maxNiceDivisors(8) == 18  # example case
assert sol.maxNiceDivisors(9) == 27  # all 3s
assert sol.maxNiceDivisors(10) == 36  # avoid remainder 1
assert sol.maxNiceDivisors(11) == 54  # 3 + 3 + 3 + 2
assert sol.maxNiceDivisors(12) == 81  # 3^4
assert sol.maxNiceDivisors(1000000000) > 0  # stress test for large input

Test Summary

Test Why
primeFactors = 1 Smallest valid input
primeFactors = 2 Base edge case
primeFactors = 3 Base edge case
primeFactors = 4 Tests 2 + 2 decomposition
primeFactors = 5 Provided example
primeFactors = 6 Pure power of 3
primeFactors = 7 Tests remainder 1 adjustment
primeFactors = 8 Provided example
primeFactors = 9 Multiple 3s
primeFactors = 10 Important remainder == 1 case
primeFactors = 11 Mixed decomposition
primeFactors = 12 Larger exact multiple of 3
primeFactors = 1000000000 Maximum scale stress test

Edge Cases

One important edge case is when primeFactors is less than or equal to 3. A common mistake is to always split numbers into smaller pieces. However, for these tiny values, splitting reduces the product rather than increasing it. For example, splitting 3 into 1 + 2 gives a product of 2, which is worse than keeping 3 itself. The implementation explicitly returns the original value for these cases.

Another critical edge case occurs when the remainder after division by 3 equals 1. A naive implementation might compute:

$$3^q \times 1$$

But multiplying by 1 wastes prime factors and produces a suboptimal product. The correct adjustment is to replace one 3 + 1 pair with 2 + 2, producing a larger multiplication result. The implementation handles this using:

pow(3, quotient - 1, MOD) * 4

A third important edge case is the extremely large upper constraint:

$$primeFactors = 10^9$$

Direct exponentiation or recursive multiplication would overflow or timeout. The implementation avoids this by using modular exponentiation, which computes powers in logarithmic time while continuously reducing intermediate values modulo 10^9 + 7.