<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:
    map<pair<int, int>, double> mp;
    vector<double> pre;

    double helper(vector<int>& nums, int k, int end){
        // cout << end << endl;
        // return(pre[end+1]);
        if(k == 1){
            return(pre[end+1] / (end+1) );
        }
        if(mp.find({end, k}) != mp.end())return(mp[{end, k}]);

        double ans = 0;
        for(int i=end-1;i>=k-2;i--){
            ans = max(ans, helper(nums, k-1, i) + (pre[end+1]-pre[i+1])/(end-i));
        }
        mp[{end, k}] = ans;
        return(ans);
    }

    double largestSumOfAverages(vector<int>& nums, int k) {
        double ans = INT_MIN;
        int n = nums.size();
        if(n == 1)return(nums[0]);

        pre.resize(n+1, 0);
        pre[0] = 0;
        for(int i=0;i<n;i++)pre[i+1] = pre[i] + nums[i];
        // for(int i=0;i<=n;i++)cout << pre[i] << " ";cout << endl;
        // return(0);
        return(helper(nums, k, n-1));
    }
};
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