<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 solve(int ind, int k, vector<int> &prefix, int n, vector<vector<double>> &dp){
      if(ind == n){
        return 0.0;
      }
      if(k == 0){
        return -1e9;
      }
      if(dp[ind][k] != -1){
        return dp[ind][k];
      }
      double res = 0;
      for(int i = ind; i < n; i++){
        // calculate the 
        int sum = prefix[i+1]-prefix[ind];
        int elements = (i-ind)+1;
        double avg = (sum*1.0)/(elements);
        double curr = avg + solve(i+1, k-1, prefix, n, dp);
        res = max(res, curr);
      }
      return dp[ind][k] = res;
    }
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> prefix(n+1, 0);
        for(int i = 0; i < n; i++){
          prefix[i+1] = prefix[i]+nums[i];
        }
        vector<vector<double>> dp(n, vector<double>(n+1, -1));
        double res = solve(0, k, prefix, n, dp);
        return res;
    }
};
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