Largest Sum of Averages
<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>
Array; Dynamic Programming; Prefix Sum
func largestSumOfAverages(nums []int, k int) float64 {
	n := len(nums)
	s := make([]int, n+1)
	for i, x := range nums {
		s[i+1] = s[i] + x
	}
	f := make([][]float64, n)
	for i := range f {
		f[i] = make([]float64, k+1)
	}
	var dfs func(int, int) float64
	dfs = func(i, k int) float64 {
		if i == n {
			return 0
		}
		if f[i][k] > 0 {
			return f[i][k]
		}
		if k == 1 {
			return float64(s[n]-s[i]) / float64(n-i)
		}
		ans := 0.0
		for j := i + 1; j < n; j++ {
			ans = math.Max(ans, float64(s[j]-s[i])/float64(j-i)+dfs(j, k-1))
		}
		f[i][k] = ans
		return ans
	}
	return dfs(0, k)
}
