<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:
    
    vector<vector<double>> dp;
    
    double help(vector<int>& nums,int k, int start)
    {
        double ans = 0;
        
        if(dp[start][k] != -1)
        {
            return dp[start][k];
        }
        
        if(k==1)
        {
            for(int i=start; i<nums.size(); i++)
            {
                ans += nums[i];
            }
            
            return dp[start][k] = ans/(nums.size()-start);
        }
        
        double sum = 0;

// Loop through the array from the start index to ensure there are enough elements left to form the remaining parts.
        for (int i = start; i <= nums.size() - k; i++) 
        {
        // Add the current element to the cumulative sum
            sum += nums[i];

            // Calculate the average of the current segment
            double average = sum / (i - start + 1);

            // Recursively calculate the maximum sum of averages for the remaining parts and add the current average
            double remainingMaxAverage = help(nums, k - 1, i + 1);

            // Update the answer with the maximum value found so far
            ans = max(ans, average + remainingMaxAverage);
        }
        
        return dp[start][k] = ans;
    }
    double largestSumOfAverages(vector<int>& nums, int k) 
    {
        
        int n = nums.size();
        
        dp.resize(101,vector<double>(101,-1));
        
        return help(nums,k,0);
    }
};

/*For the given array [9, 1, 2, 3, 9] and k = 3, we try various partitions:
[9], [1, 2, 3], [9]
[9, 1], [2], [3, 9]
[9, 1, 2], [3], [9]
And so on...*/
15400