<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 {
private:
    double solve(vector<vector<double>>& section, int k, map<pair<pair<int,int>,int>,double>& mp, int left, int right){
        if (k == 1) return section[left][right];
        if (k != 1 && left == right) return INT_MIN;
        if (mp.count({{left, right}, k})) return mp[{{left, right}, k}];
        double ans = 0;
        for (int i=left;i<right;++i){
            ans = max(ans, section[left][i] + solve(section, k-1, mp, i+1, right));
        }
        return mp[{{left, right}, k}] = ans;
    }
public:
    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        vector<vector<double>> section(n, vector<double>(n));
        for (int i=0;i<nums.size();++i){
            double sum = 0;
            for (int j=i;j<nums.size();++j){
                sum += (double) nums[j];
                section[i][j] = sum / (double) (j-i+1);
            }
        }
        map<pair<pair<int,int>,int>,double> mp;
        return solve(section, k, mp, 0, n-1);
    }
};
15000