<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]:
        vals = self.get_vals(nums, k)
        #print(vals)
        l, r = self.rec(nums, 3, 0, k, vals, dict())
        #print(l,r)
        return l
    def rec(self, nums, left, i, k, vals, d):
        key = i, left
        if key in d: return d[key]

        n = len(nums)
        if left == 0 or i not in vals: return [], 0
        lst_take, res_take = self.rec(nums, left-1, i+k, k, vals, d)
        res_take += vals[i]
        lst, res = self.rec(nums, left, i+1, k, vals, d)
        if res_take >= res:
            res = res_take
            lst = [i]+lst_take
        d[key] = (lst, res)
        return d[key]
    def get_vals(self, nums, k):
        d = dict()
        n = len(nums)
        d[0] = sum(nums[:k])
        for end in range(k, n):
            d[end-k+1] = d[end-k]-nums[end-k]+nums[end]
        return d
34000