<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>

from sortedcontainers import SortedList
class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        n = len(nums)
        windows = [0]
        for i in range(k):
            windows[-1] += nums[i]
        
        for i in range(k,n):
            windows.append(windows[-1]-nums[i-k]+nums[i])
        
        pos = SortedList()
        left = SortedList()
        right = SortedList()

        N = len(windows)
        
        for i in range(2*k,N):
            right.add((-windows[i],i))
        
        for i in range(k,N-k):
            left.add((-windows[i-k],i-k))
            A,B = left[0]
            C,D = right[0]
            pos.add((-windows[i]+A+C,B,i,D))
            right.remove((-windows[i+k],i+k))
        
        _,A,B,C=pos[0]

        
        return [A,B,C]
23000