<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:
        # We must partition to k groups, it is always beneficial!

        # Find all subarray sums and averages
        prefix_sum = [0]
        for n in nums:
            prefix_sum.append(prefix_sum[-1]+n)
        memo = {}
        for i in range(len(nums)):
            for j in range(i, len(nums)):
                cur_sum = prefix_sum[j+1] - prefix_sum[i]
                cur_len = j-i+1
                cur_avg = cur_sum / cur_len
                memo[i, j] = cur_avg
        
        # Search for all partitions
        helper_memo = {}
        def helper(cur_pos, k):
            if (cur_pos, k) in helper_memo:
                return helper_memo[cur_pos, k]
            if k == 1:
                return memo[cur_pos, len(nums)-1]
            max_score = 0
            for partition_pos in range(cur_pos, len(nums)-k+1):
                cur_score = memo[cur_pos, partition_pos] + helper(partition_pos+1, k-1)
                max_score = max(max_score, cur_score)
            helper_memo[cur_pos, k] = max_score
            return max_score

        return helper(0, k)
20600