Naive Solution: Brute Force

A brute-force approach would involve trying all possible combinations of positive integers that sum up to n and then calculating the product of each combination. We would then keep track of the maximum product found so far.

This approach is highly inefficient because the number of combinations grows exponentially with n.

Big O Runtime: O(k^n), where k is a small constant Big O Space: O(n), due to recursion depth or storing the combinations.

Optimal Solution: Dynamic Programming or Mathematical Approach

A more efficient solution leverages dynamic programming or a mathematical observation.

Mathematical Approach

The key insight is that to maximize the product, we should break the number n into as many 3s as possible. If n is divisible by 3, then the maximum product is simply 3^(n/3). If n has a remainder of 1 when divided by 3, then replace the last 3 with a 4 (2 * 2). If n has a remainder of 2 when divided by 3, then multiply the rest by 2.

Let's prove why splitting into 3's is optimal:

• Splitting into 1's is obviously bad because the product will always be 1.
• Consider splitting into 2's. We know that 2 * 2 * 2 = 8, and 3 * 3 = 9. So replacing three 2's with two 3's will increase the product.
• Consider the case where the numbers are greater than 3. For example, consider breaking 5 into 2 + 3 vs 5. We know that 2 * 3 = 6 > 5. So we can keep breaking the numbers until they become 2 or 3.

Edge Cases:

• n = 2, the result is 1 (1 * 1)
• n = 3, the result is 2 (1 * 2)

Code Implementation (Python):

def integerBreak(n: int) -> int:
    if n == 2:
        return 1
    if n == 3:
        return 2

    num_threes = n // 3
    remainder = n % 3

    if remainder == 0:
        return 3 ** num_threes
    elif remainder == 1:
        return (3 ** (num_threes - 1)) * 4
    else:
        return (3 ** num_threes) * 2

Big O Runtime: O(1) Big O Space: O(1)

Dynamic Programming Approach

We can also solve this using dynamic programming. Let dp[i] be the maximum product we can get by breaking the integer i.

The base cases are dp[1] = 1.

The transition function is: dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j])) for all 1 <= j < i

Code Implementation (Python):

def integerBreak_dp(n: int) -> int:
    dp = [0] * (n + 1)
    dp[1] = 1

    for i in range(2, n + 1):
        for j in range(1, i):
            dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j]))

    return dp[n]

Big O Runtime: O(n^2) Big O Space: O(n)