<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:
    double largestSumOfAverages(vector<int>& nums, int k) {
        int prefixSum = 0;
        int numItems = nums.size();
        vector<int> prefixSums;
        for (int i = 0; i < numItems; i++) {
            prefixSum += nums[i];
            prefixSums.push_back(prefixSum);
        }

        
        vector<vector<vector<double>>> dynamic = vector<vector<vector<double>>>(k, vector<vector<double>>(numItems, vector<double>(numItems, 0)));
        for (int len = 1; len <= k; len++) {
            for (int start = 0; start <= numItems - len; start++) {
                for (int end = start + len - 1; end < numItems; end++) {   
                    if (len == 1) {
                        int prefixSum = prefixSums[end] - ((start > 0) ? prefixSums[start - 1] : 0);
                        double avg = 1.0 * prefixSum / (end - start + 1);
                        dynamic[len - 1][start][end] = avg;
                    } else {
                        double maxPartition = 0;
                        for (int partitionEnd = start; partitionEnd < end - len + 2; partitionEnd++) {
                            //cout << "start is " << start << ", partitionEnd is: " << partitionEnd << ", len is " << len << "\n";
                            int prefixSum = prefixSums[partitionEnd] - ((start > 0) ? prefixSums[start - 1] : 0);
                            double avg = 1.0 * prefixSum / (partitionEnd - start + 1);
                            double partition = avg + dynamic[len - 2][partitionEnd + 1][end];
                            maxPartition = max(maxPartition, partition);
                        }
                        dynamic[len - 1][start][end] = maxPartition;

                    }

//                    cout << "dynamic[" << len - 1 << "][" << start << "][" << end << "] is " << dynamic[len - 1][start][end] << "\n";
                }
            }
        }
        return dynamic[k - 1][0][numItems - 1];
    }
};
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