<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]:
        l = [0] * len(nums)
        curr = sum(nums[:k])
        l[k-1] = (curr, 0)
        for i in range(k, len(nums)):
            curr += nums[i] - nums[i-k]
            if l[i-1][0] >= curr:
                l[i] = l[i-1]
            else:
                l[i] = (curr, i-k+1)
        
        curr = sum(nums[len(nums)-k:len(nums)])
        curr_max = curr
        curr_max_start = len(nums)-k
        mid_group = sum(nums[len(nums)-(2*k):len(nums)-k])
        res_value = curr + mid_group + l[len(nums)-(2*k)-1][0]
        res = [l[len(nums)-(2*k)-1][1], len(nums)-(2*k), len(nums)-k]
        for i in range(len(nums)-k-1, 2*k - 1, -1):
            mid_group += nums[i-k] - nums[i]
            curr += nums[i] - nums[i+k]
            if curr >= curr_max:
                curr_max = curr
                curr_max_start = i
            curr_max = max(curr_max, curr)
            curr_triple_value = l[i-k-1][0] + mid_group + curr_max
            if curr_triple_value >= res_value:
                res_value = curr_triple_value
                res = [l[i-k-1][1], i-k, curr_max_start]
            
        return res
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