<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 {
  
    double func(int ind,int k,vector<int>&arr,vector<vector<double>>&dp){
        int n = arr.size();
        if(k==0){
            double ct = 0 , sum = 0;
            for(int i = ind;i<n;i++){
                sum +=arr[i];
                ct++;
            }
            return (sum/ct);
        }
        if(ind>=(n-1)) return -1e8;
        if(dp[ind][k]!=-1) return dp[ind][k];
        double ans = 0 , sum = 0;
        for(int i = ind;i<(n-1);i++){
            sum += arr[i];
            double val = (sum/(i-ind+1)) + func(i+1 , k-1 , arr , dp);
            ans = max(ans , val);
        }
        return dp[ind][k] = ans;
    }
public:
    double largestSumOfAverages(vector<int>& nums, int k) {
        k--;
        double cc = -1;
       vector<vector<double>>dp(101 , vector<double>(101 , cc));
        return func(0 , k , nums , dp);
    }
};
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