<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)
        dk, tk = 2*k, 3*k
        maxOneSub = [[[], 0] for _ in range(N)]
        maxTwoSub = [[[], 0] for _ in range(N)]
        maxThreeSub = [[[], 0] for _ in range(N)]
        maxOneSub[k-1] = [[0], sum(nums[:k])]
        maxTwoSub[dk-1] = [[0, k], maxOneSub[k-1][1]+sum(nums[k:dk])]
        maxThreeSub[tk-1] = [[0, k, dk], maxTwoSub[dk-1][1]+sum(nums[dk:tk])]
        
        for i in range(k, N):
            curksum = sum(nums[i-k+1: i+1])
            if maxOneSub[i-1][1] >= curksum:
                maxOneSub[i] = maxOneSub[i-1].copy()
            else:
                maxOneSub[i] = [[i-k+1], curksum]
            if i >= dk:
                cur2ksum = maxOneSub[i-k][1]+curksum
                if maxTwoSub[i-1][1] >= cur2ksum:
                    maxTwoSub[i] = maxTwoSub[i-1].copy()
                else:
                    maxTwoSub[i] = [maxOneSub[i-k][0]+[i-k+1], cur2ksum]
            if i >= tk:
                cur3ksum = maxTwoSub[i-k][1]+curksum
                if maxThreeSub[i-1][1] >= cur3ksum:
                    maxThreeSub[i] = maxThreeSub[i-1].copy()
                else:
                    maxThreeSub[i] = [maxTwoSub[i-k][0]+[i-k+1], cur3ksum]
                    
        return maxThreeSub[N-1][0]
        
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