<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]:
        W = [] #array of sums of windows
        curr_sum = 0
        for i, x in enumerate(nums):
            curr_sum += x
            if i >= k: 
                curr_sum -= nums[i - k]
            if i >= k - 1: 
                W.append(curr_sum)

        left = [0] * len(W)
        best = 0
        for i in range(len(W)):
            if W[i] > W[best]:
                best = i
            left[i] = best

        right = [0] * len(W)
        best = len(W) - 1
        for i in range(len(W) - 1, -1, -1):
            if W[i] >= W[best]:
                best = i
            right[i] = best

        ans = None
        for j in range(k, len(W) - k):
            i, l = left[j - k], right[j + k]
            if ans is None or (W[i] + W[j] + W[l] > W[ans[0]] + W[ans[1]] + W[ans[2]]):
                ans = i, j, l
        return ans

        
19800