<p>Given an integer <code>n</code>, break it into the sum of <code>k</code> <strong>positive integers</strong>, where <code>k &gt;= 2</code>, and maximize the product of those integers.</p>

<p>Return <em>the maximum product you can get</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> n = 2
<strong>Output:</strong> 1
<strong>Explanation:</strong> 2 = 1 + 1, 1 &times; 1 = 1.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> n = 10
<strong>Output:</strong> 36
<strong>Explanation:</strong> 10 = 3 + 3 + 4, 3 &times; 3 &times; 4 = 36.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>2 &lt;= n &lt;= 58</code></li>
</ul>

class Solution:
    def integerBreak(self, n: int) -> int:
        memo = {1 : 1}
        def dp(s):
            if s in memo:
                return memo[s]
        
            res = 0 if s == n else s

            for j in range(1,s):
                a = dp(j) * dp(s-j)
                res = max(res,a)
            memo[s] = res
            return res
        return dp(n)







            


            

            
            

        
16600