<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>

from typing import List
from math import inf

class Solution:
    def largestSumOfAverages(self, nums: List[int], k: int) -> float:
        prefix = [0] * (len(nums) + 1)

        # Building prefix sum array
        for i in range(1, len(nums) + 1):
            prefix[i] = prefix[i - 1] + nums[i - 1]

        # Memoization to optimize repeated calls
        memo = {}

        def dfs(i, k):
            if (i, k) in memo:
                return memo[(i, k)]
            if i >= len(nums) :return -inf
            if k == 1:
                return (prefix[-1] - prefix[i]) / (len(nums) - i)

            result = -inf
            for j in range(len(nums), i , -1):
                result = max(result, dfs(j, k - 1) + (prefix[j] - prefix[i]) / (j - i))

            memo[(i, k)] = result
            return result

        return dfs(0, k)

18100