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
function largestSumOfAverages(nums: number[], k: number): number {
    const n = nums.length;
    const s: number[] = Array(n + 1).fill(0);
    for (let i = 0; i < n; i++) {
        s[i + 1] = s[i] + nums[i];
    }
    const f: number[][] = Array.from({ length: n }, () => Array(k + 1).fill(0));
    const dfs = (i: number, k: number): number => {
        if (i === n) {
            return 0;
        }
        if (f[i][k] > 0) {
            return f[i][k];
        }
        if (k === 1) {
            return (s[n] - s[i]) / (n - i);
        }
        for (let j = i + 1; j < n; j++) {
            f[i][k] = Math.max(f[i][k], dfs(j, k - 1) + (s[j] - s[i]) / (j - i));
        }
        return f[i][k];
    };
    return dfs(0, k);
}
