Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

Example 1:

Input: 2
Output: 1
Explanation: 2 = 1 + 1, 1 × 1 = 1.

Example 2:

Input: 10
Output: 36
Explanation: 10 = 3 + 3 + 4, 3 × 3 × 4 = 36.

Note: You may assume that n is not less than 2 and not larger than 58.

 

Approach #1: Math. [C++]

class Solution {
public:
    int integerBreak(int n) {
        static vector<int> arr(7, 0);
        arr[2] = 1, arr[3] = 2;
        arr[4] = 4, arr[5] = 6, arr[6] = 9;
        int ans = 1;
        
        while (n > 6) 
            n -= 3, ans *= 3;
        
        return ans * arr[n];
    }
};

　　

Analysis:

For convenience, say n is sufficiently large and can be broken into any smaller real positive numbers. We now try to calculate which real number generates the largest product.
Assume we break n into (n / x) x's, then the product will be xn/x, and we want to maximize it.

 

Taking its derivative gives us n * xn/x-2 * (1 - ln(x)).
The derivative is positive when 0 < x < e, and equal to 0 when x = e, then becomes negative when x > e,
which indicates that the product increases as x increases, then reaches its maximum when x = e, then starts dropping.

 

This reveals the fact that if n is sufficiently large and we are allowed to break n into real numbers,
the best idea is to break it into nearly all e's.
On the other hand, if n is sufficiently large and we can only break n into integers, we should choose integers that are closer to e.
The only potential candidates are 2 and 3 since 2 < e < 3, but we will generally prefer 3 to 2. Why?

 

Of course, one can prove it based on the formula above, but there is a more natural way shown as follows.

 

6 = 2 + 2 + 2 = 3 + 3. But 2 * 2 * 2 < 3 * 3.
Therefore, if there are three 2's in the decomposition, we can replace them by two 3's to gain a larger product.

 

All the analysis above assumes n is significantly large. When n is small (say n <= 10), it may contain flaws.
For instance, when n = 4, we have 2 * 2 > 3 * 1.
To fix it, we keep breaking n into 3's until n gets smaller than 10, then solve the problem by brute-force.

 

Reference:

https://leetcode.com/problems/integer-break/discuss/80721/Why-factor-2-or-3-The-math-behind-this-problem.