<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. You can partition the array into <strong>at most</strong> <code>k</code> non-empty adjacent subarrays. The <strong>score</strong> of a partition is the sum of the averages of each subarray.</p>

<p>Note that the partition must use every integer in <code>nums</code>, and that the score is not necessarily an integer.</p>

<p>Return <em>the maximum <strong>score</strong> you can achieve of all the possible partitions</em>. Answers within <code>10<sup>-6</sup></code> of the actual answer will be accepted.</p>

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

<pre>
<strong>Input:</strong> nums = [9,1,2,3,9], k = 3
<strong>Output:</strong> 20.00000
<strong>Explanation:</strong> 
The best choice is to partition nums into [9], [1, 2, 3], [9]. The answer is 9 + (1 + 2 + 3) / 3 + 9 = 20.
We could have also partitioned nums into [9, 1], [2], [3, 9], for example.
That partition would lead to a score of 5 + 2 + 6 = 13, which is worse.
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6,7], k = 4
<strong>Output:</strong> 20.50000
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 100</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= k &lt;= nums.length</code></li>
</ul>

class Solution {
public:
    // int memo[]
    // returns avg, partitions
    double memo[101][101][101];
    double rec(int i, int j, int k, vector<double>& pref) {
        if (j == pref.size() -1) return (pref[j] - pref[i]) / (j - i);
        if (memo[i][j][k] != -1) return memo[i][j][k];
        auto ans = rec(i, j + 1, k, pref);
        if (k > 0 && i != j) {
            auto ans2 = rec(j, j + 1, k - 1, pref);
            ans = max(ans, (pref[j] - pref[i]) / (j - i) + ans2);
        }
        return memo[i][j][k] = ans;
    }

    double largestSumOfAverages(vector<int>& nums, int k) {
       // return maximum possible sum of averages of subsets
       vector<double> pref(nums.size() + 1, 0);
       for (int i =0; i < 101; i++) {
        for (int j=0; j < 101; j++) {
            for (int k =0; k < 101; k++) memo[i][j][k] = -1;
        }
       }
       for (int i =0; i < nums.size(); i++) pref[i + 1] = pref[i] + nums[i]; 
        return rec(0, 0, k -1, pref);
    }
};
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