<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]:
        """
        requirements:
        - is it guarantee that len(nums) >= 3 * k? Yes

        solution:
        - dfs(idx, remain) -> returns max sum of remain number of k length subarrays starting idx
        - for this question we can return dfs(0, 3)

        dfs(idx, remain) = max(dfs(idx + 1, remain), dfs(idx + k, remain - 1))

        - add lru_cache to dfs
        """
        @lru_cache()
        def dfs(idx, remain):
            if remain == 0:
                return 0, []
            if len(nums) - idx < remain * k:
                return 0, []
            sum1, indexes1 = dfs(idx + 1, remain)
            sum2, indexes2 = dfs(idx + k, remain - 1)
            sum2 += sum(nums[idx:idx + k])
            indexes2 = [idx] + indexes2
            if sum1 > sum2:
                return sum1, indexes1
            return sum2, indexes2

        return dfs(0, 3)[1]
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