<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]:
        s,cumsum=0,[0]
        for n in nums:
            s+=n
            cumsum.append(s)
        dp=[[(0,[]) for _ in range(4)] for _ in range(len(nums)+1)]
        for i in range(k,len(nums)+1):
            for arrs in range(1,4):
                skip = dp[i-1][arrs][0]
                use = dp[i-k][arrs-1][0]+cumsum[i]-cumsum[i-k] if i>=k else 0
                if skip>=use:
                    dp[i][arrs]=(skip, dp[i-1][arrs][1])
                else:
                    incl_i=dp[i-k][arrs-1][1].copy()
                    incl_i.append(i-k)
                    dp[i][arrs]=(use, incl_i)
        return dp[len(nums)][3][1]
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