<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 f(int n , int i , vector<vector<int>>&dp){
        if( n == 0 ) return 1;
        if( i == 0 ) return 0;

        if(dp[i][n]!=-1) return dp[i][n];

        long long np = f(n , i - 1 , dp);

        long long p = 0;
        if(i<=n){
            p = i*f(n - i, i , dp);
        }

        return dp[i][n] = max(p , np);
    }

    int integerBreak(int n) {
        vector<vector<int>>dp(n , vector<int>(n+1 , -1));

        return f(n , n-1 , dp);
    }
};
8100