<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:
    def largestSumOfAverages(self, nums: List[int], k: int) -> float:
        self.nums = nums 
        self.k = k 
        self.cache = {} 

        return self.largest_partition(0, 0)


    def largest_partition(self, i: int, partition: int) -> float: 
        if i == len(self.nums): 
            return 0

        if partition >= self.k: 
            return -math.inf

        if (i, partition) in self.cache: 
            return self.cache[(i, partition)]

        # start moving 

        largest_partition_average = -math.inf

        running_sum = 0 
        for j in range(i, len(self.nums)):
            size = j - i + 1
            running_sum += self.nums[j]
            new_partition_average = (running_sum / size) + self.largest_partition(j + 1, partition + 1)
            largest_partition_average = max(largest_partition_average, new_partition_average)

        self.cache[(i, partition)] = largest_partition_average
        return largest_partition_average


        
17100