<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]:
        pfxs = [0] + list(accumulate(nums))
        dp = [[[] for k in range(3)] for i in range(len(nums))]
        def solve(idx, n):
            if n==0 or idx >= len(nums)-k+1: return 0, []
            if dp[idx][n-1] != []: return dp[idx][n-1]
            if pfxs[idx+k]-pfxs[idx] + solve(idx+k, n-1)[0] >= solve(idx+1, n)[0]:
                mx = pfxs[idx+k]-pfxs[idx] + solve(idx+k, n-1)[0]
                res = [idx] + solve(idx+k, n-1)[1]
            else:
                mx, res = solve(idx+1, n)[0], solve(idx+1, n)[1]
            dp[idx][n-1] = [mx, res]
            return mx, res
        return solve(0, 3)[1]
36700