<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]:
        result = []
        current_sum = 0
        memo = {}
        for i, num in enumerate(nums):
            current_sum += num
            nums[i] = current_sum

        def helper(idx, used, k):
            if used == 3:
                return 0, []
            
            if idx + k -1 >=len(nums):
                return 0, []

            if (idx, used) in memo:
                return memo[(idx, used)]
            
            take_sum, take_indicies = helper(idx+k, used+1, k)
            take_sum += nums[idx+k-1] if idx == 0 else nums[idx+k-1] - nums[idx-1] 
            dont_take_sum, dont_take_indicies = helper(idx+1, used, k)

            if take_sum >= dont_take_sum:
                memo[(idx, used)] = (take_sum, [idx] + take_indicies)
            else:
                memo[(idx, used)] = (dont_take_sum, dont_take_indicies)
            return memo[(idx, used)]
        result = helper(0, 0, k)
        return result[1]
38000