<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 dp[101][101][101]; 

double solve(int i, int j, int parts, vector<int>& prefx, int k, int n) {
    if (j>n) return 0; 
    if(parts==k) return INT_MIN;

    double &ret = dp[i][j][parts];
    if (ret != -1.0) return ret;  

    double avg = (double)(prefx[j] - prefx[i - 1]) / (j - (i - 1));

    ret = (double)max(avg + solve(j + 1, j + 1, parts + 1, prefx, k, n), 
              solve(i, j + 1, parts, prefx, k, n));

    return ret;
}

double largestSumOfAverages(vector<int>& nums, int k) {
    for (int i = 0; i < 101; ++i) {
        for (int j = 0; j < 101; ++j) {
            for (int k = 0; k < 101; ++k) {
                dp[i][j][k] = -1.0;
            }
        }
    }
    
    int n = nums.size();
    vector<int> prefx(n + 1, 0);
    for (int i = 1; i <= n; ++i) {
        prefx[i] = nums[i - 1] + prefx[i - 1];
    }
    return solve(1, 1, 0, prefx, k, n);
}
};
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