<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>

def max_sum(nums, curr, k, subs, cache, ksum_at):
    remaining_elements = len(nums) - curr
    assert remaining_elements >= k * subs

    idx = (curr, subs)
    if idx not in cache:
        best_sum = None
        best_sol = None
        for left in reversed(range(curr, len(nums) - k * subs + 1)):
            # print('Gnerating for', left, subs)
            sub_sols = max_sum(nums, left + k, k, subs - 1, cache, ksum_at)
            # print('sub_sols for', left, subs, sub_sols)
            curr_sol = [left] + sub_sols
            curr_sum = sum([ksum_at[sol_idx] for sol_idx in curr_sol])
            if best_sum is None or curr_sum >= best_sum:
                best_sum = curr_sum
                best_sol = curr_sol

            cache[(left, subs)] = best_sol

    return cache[idx]


class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        cache = {}
        ksum_at = []
        window_sum = sum(nums[:k])
        ksum_at.append(window_sum)

        for left in range(1, len(nums) - k + 1):
            window_sum -= nums[left - 1]
            window_sum += nums[left + k - 1]
            ksum_at.append(window_sum)

        for i in reversed(range(len(nums) - k + 1)):
            if (i + 1, 1) not in cache or ksum_at[cache[(i + 1, 1)][0]] <= ksum_at[i]:
                cache[(i, 1)] = [i]
            else:
                cache[(i, 1)] = cache[(i + 1, 1)]

        # print(ksum_at)
        # print(cache)
        for i in range(1, 4):
            max_sum(nums, 0, k, i, cache, ksum_at)
        # print(cache)
        return max_sum(nums, 0, k, 3, cache, ksum_at)
27700