<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> m;
    double f(int i, vector<int> &nums, int k)
    {
        int n=nums.size();
        if(i!=n && k==0) return INT_MIN;
        if(i==n || k<=0) return 0;
        if(m.find({i,k})!=m.end()) return m[{i,k}];
        double sum=0, count=0, ans=0;
        for(int x=i; x<=n-k; x++)
        {
            count++;
            sum+=nums[x];
            ans=max(ans,sum/count+f(x+1,nums,k-1));
        }
        return m[{i,k}]=ans;
    }
    double largestSumOfAverages(vector<int>& nums, int k) 
    {
        double mx=0;
        for(int x=1; x<=k; x++)
        {
            mx=max(mx,f(0,nums,x));
        }
        return mx;
    }
};
13200