<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:
        # more partitions always gets larger score
        if k == len(nums):
            return float(sum(nums))
        # make prefix sum array
        pref = [0]
        s = 0
        for n in nums:
            s += n
            pref.append(s)
        # kth best = max([nums[0,k-1] ~ nums[0,-2] best partition(k-1) + avg(remaining)])
        # make dp table of dp[k][n] for best score of k partion of nums[:n]
        dp = [[0]*len(nums) for _ in range(k+1)]
        # k = 1 is just avg from each prefix
        for j in range(len(nums)):
            dp[1][j] = pref[j+1] / (j + 1)
        for i in range(2, k+1):
            prev, cur = dp[i-1], dp[i]
            for j in range(i-1, len(nums)):
                opt = 0
                for l in range(j):
                    opt = max(opt, prev[l] + (pref[j+1] - pref[l+1]) / (j-l))
                cur[j] = opt
        return dp[k][-1]
16700