<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>& nums, int k) {
        int n = nums.size();
        vector<int> preSum(n + 1, 0);
        for (int i = 0; i < n; i++) preSum[i + 1] = preSum[i] + nums[i];
        // vector<vector<double>>dp(n+1,vector<double>(k+1,-1.0));
        map<pair<int,int>,double>mp;
        function<double(int, int)> solve = [&](int ind, int cnt) -> double {
            if (cnt == 1) 
            {
                return preSum[ind + 1] * 1.0 / (ind + 1);
            }
            // if(dp[ind][cnt]!=-1.0) return dp[ind][cnt];
            if(mp.find({ind,cnt})!=mp.end()) return mp[{ind,cnt}];
            double ans = 0.0;
            for (int i = 1; i <= ind + 1; i++) 
            {
                double sum = preSum[ind + 1] - preSum[ind - i + 1];
                double avg = sum / i;
                double rtn = solve(ind - i, cnt - 1);
                ans = max(ans, avg + rtn);
            }

            return mp[{ind,cnt}] = ans;
        };
        
        return solve(n - 1, k);
    }
};

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