<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]:
        memo = {}
        res = self.window_sum(nums, k, 0, 0, memo)
        return res[1]
    
    def window_sum(self, nums, k, idx, used, memo):
        # we can only have 3 windows at a time
        if used == 3:
            return 0, []
        
        # are we going to overflow over our nums array?
        if idx - (used * k) > len(nums):
            return 0, []
        
        if (idx, used) in memo:
            return memo[(idx, used)]
        
        take_curr_sum, take_curr_indices = self.window_sum(nums, k, idx + k, used + 1, memo)
        take_curr_sum += sum(nums[idx: idx + k])

        skip_curr_sum, skip_curr_indices = self.window_sum(nums, k, idx + 1, used, memo)

        if take_curr_sum >= skip_curr_sum:
            memo[(idx, used)] = (take_curr_sum, ([idx] + take_curr_indices))
            return memo[(idx, used)]
        else:
            memo[(idx, used)] = (skip_curr_sum, skip_curr_indices)
            return memo[(idx, used)]
36800