<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 helper(int ind, int pleft, vector<int> &pfs, vector<int> &nums, vector<vector<double>> &dp){

        if(ind>=pfs.size()){
            if(pleft>0) return -1e9; // i still didnt do some partitions which is wrong -> penalise
            else return 0;
        }

        if(dp[ind][pleft]!=-1) return dp[ind][pleft];

        if( pleft==0 ){ // absolutely crt
            // return remaining sum avg
            double sum = 0;
            if(ind!=0) sum = pfs.back() - pfs[ind-1];
            else sum = pfs.back();
            double len = (pfs.size() - 1) -  ind + 1;

            return sum/len;
        }

        double maxavg = 0;

        for(int cut = ind; cut <=nums.size()-2 ; cut++ ){
                double sum = 0;
                if(ind!=0) sum = pfs[cut] - pfs[ind-1];
                else sum = pfs[cut];

                double len = cut-ind+1;

                double curans = sum/len + helper(cut+1,pleft-1,pfs,nums,dp);
                
                maxavg = max(maxavg, curans);
        }

        return dp[ind][pleft] = maxavg;
    }


    double largestSumOfAverages(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> pfs(n);

        for(int i=0; i<n; i++){
            if(i==0) pfs[i] = nums[i];
            else pfs[i] = pfs[i-1] + nums[i];
        }

        vector<vector<double>> dp(n+1, vector<double>(k+1,-1));

        return helper(0,k-1,pfs,nums,dp);
    }
};
10000