<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>& A, int k) {
        int n = A.size();
        vector<int> presum(n+1, 0);
        for (int i = 1; i <= n; i++) presum[i] = presum[i-1] + A[i-1];
        unordered_map<string,double> memo;
        return helper(A, k, 0, n-1, presum, memo);   
    }

    double helper(vector<int> &A, int k, int left, int right, vector<int> &presum, unordered_map<string,double> &memo) {
        if (k == 1) {
            return (double) (presum[right+1] - presum[left]) / (right - left + 1);
        }
        string key = to_string(left) + '-' + to_string(right) + '-' + to_string(k);
        if (memo.count(key)) return memo[key];
        double res = helper(A, 1, left, right, presum, memo);
        for (int i = left; i < right; i++) {
            double cur = helper(A, 1, left, i, presum, memo) + 
                         helper(A, k-1, i+1, right, presum, memo);
            res = max(res, cur);
        }
        memo[key] = res;
        return res;
    }
};
16000