<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>

import itertools as it
class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        psums = list(it.accumulate(nums))
        bests = [[(0, None), (0, None), (0, None)] for _ in nums]
        bests[k - 1][0] = (psums[k - 1], (0,))
        
        for i in range(k, len(nums)):
            v = psums[i] - psums[i - k]
            if v > bests[i - 1][0][0]:
                bests[i][0] = (v, (i - k + 1,))
            else:
                bests[i][0] = bests[i - 1][0]
            for r in range(1, 3):
                if bests[i - k][r - 1][1] is not None and v + bests[i - k][r - 1][0] > bests[i - 1][r][0]:
                    bests[i][r] = (v + bests[i - k][r - 1][0], bests[i - k][r - 1][1] + (i - k + 1,))
                else:
                    bests[i][r] = bests[i - 1][r]
        return bests[-1][-1][1]

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