<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 {
public:
    int integerBreak(int n) {
        vector<int> dp(n+1,0);
        if(n == 2) return 1;
        if(n == 3) return 2;
        dp[1] = 1;
        dp[2] = 2; // not correct actually
        dp[3] = 3; // not correct actually

        for(int i = 4; i <= n; i++){
            for(int j = 1; j < i ; j++){
                dp[i] = max(dp[i], dp[i-j] * dp[j]);
            }
        }
        return dp[n];
    }
};
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