<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]:
        n = len(nums)
        sums = [0] * (n - k + 1)
        for i in range(n - k + 1):
            sums[i] = sum(nums[i:i+k])
        
        left = [0] * len(sums)
        right = [0] * len(sums)

        best = 0
        for i in range(len(sums)):
            if sums[i] > sums[best]:
                best = i
            left[i] = best
        
        best = len(sums)-1
        for i in range(len(sums)-1, -1, -1):
            if sums[i] >= sums[best]:
                best = i
            right[i] = best
        
        res = []
        max_sum = 0
        for j in range(k, len(sums)-k):
            i, l = left[j-k], right[j+k]
            curr_sum = sums[i] + sums[j] + sums[l]
            if curr_sum > max_sum:
                max_sum = curr_sum
                res = [i, j, l]
        
        return res
20000