<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)
        prefix = [0] * (n-(k-1))

        l = 0
        count = 0
        for i in range(n):
            count += nums[i]

            if (i+1)-l > k:
                count -= nums[l]
                l += 1

            if (i+1)-l == k:
                prefix[i-(k-1)] = count

        sub1 = [[0, 0] for i in range(n-(k-1))]
        sub2 = [[0, 0, 0] for i in range(n-(k-1))]
        sub3 = [[0, 0, 0, 0] for i in range(n-(k-1))]

        for i in range(n-(k-1)):
            if sub1[i-1][0] < prefix[i]:
                sub1[i][0] = prefix[i]
                sub1[i][1] = i
            else:
                sub1[i] = list(sub1[i-1])

            if i >= k and sub2[i-1][0] < sub1[i-k][0] + prefix[i]:
                sub2[i][0] = sub1[i-k][0] + prefix[i]
                sub2[i][1] = sub1[i-k][1]
                sub2[i][2] = i
            else:
                sub2[i] = list(sub2[i-1])

            if i >= (k*2) and sub3[i-1][0] < sub2[i-k][0] + prefix[i]:
                sub3[i][0] = sub2[i-k][0] + prefix[i]
                sub3[i][1] = sub2[i-k][1]
                sub3[i][2] = sub2[i-k][2]
                sub3[i][3] = i
            else:
                sub3[i] = list(sub3[i-1])

        return sub3[-1][1:]
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