<p>Given an integer array <code>nums</code> and an integer <code>k</code>, find three non-overlapping subarrays of length <code>k</code> with maximum sum and return them.</p>

<p>Return the result as a list of indices representing the starting position of each interval (<strong>0-indexed</strong>). If there are multiple answers, return the lexicographically smallest one.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,2,6,7,5,1], k = 2
<strong>Output:</strong> [0,3,5]
<strong>Explanation:</strong> Subarrays [1, 2], [2, 6], [7, 5] correspond to the starting indices [0, 3, 5].
We could have also taken [2, 1], but an answer of [1, 3, 5] would be lexicographically larger.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,1,2,1,2,1,2,1], k = 2
<strong>Output:</strong> [0,2,4]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;&nbsp;2<sup>16</sup></code></li>
	<li><code>1 &lt;= k &lt;= floor(nums.length / 3)</code></li>
</ul>

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:

        p = [0] + list(accumulate(nums))
        a = [(u, u + k - 1, (p[u + k] - p[u])) for u in range(0, len(nums) - k + 1)]

        n, d = len(a), 3

        dp = [[-inf for _ in range(d + 1)] for _ in range(n + 1)]
        for i in range(n + 1): dp[i][0] = 0

        h = defaultdict(list)
        for i in range(1, n + 1):
            for dd in range(1, d + 1):
                dp[i][dd] = max(dp[i][dd], dp[i - 1][dd])
                pv = (dp[i - k][dd - 1] if i - k >= 0 else 0)
                if dp[i][dd] < a[i - 1][2] + pv:
                    dp[i][dd] = a[i - 1][2] + pv
                    h[dp[i][dd]] = h[dp[i - k][dd - 1]] + [a[i - 1][0]]

        return h[dp[n][dd]]
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