<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]:
        n = len(nums)
        s = list(accumulate(nums, initial=0))
        pre = [[] for _ in range(n)]
        suf = [[] for _ in range(n)]
        t = idx = 0
        for i in range(n - k + 1):
            if (cur := s[i + k] - s[i]) > t:
                pre[i + k - 1] = [cur, i]
                t, idx = pre[i + k - 1]
            else:
                pre[i + k - 1] = [t, idx]
        t = idx = 0
        for i in range(n - k, -1, -1):
            if (cur := s[i + k] - s[i]) >= t:
                suf[i] = [cur, i]
                t, idx = suf[i]
            else:
                suf[i] = [t, idx]
        t = 0
        ans = []
        for i in range(k, n - 2 * k + 1):
            if (cur := s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0]) > t:
                ans = [pre[i - 1][1], i, suf[i + k][1]]
                t = cur
        return ans
22800