<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, k):
        
        sm1 = sm2 = sm3 = 0
        
        acc = list(accumulate(nums, initial = 0))

        for i, (a0,a1,a2,a3) in enumerate(zip(acc      ,
                                              acc[k:]  , 
                                              acc[2*k:], 
                                              acc[3*k:])):
            if a1 - a0 > sm1:
                sm1, idx1 = a1 - a0, i

            if a2 - a1 > sm2 - sm1:
                sm2, idx2 = sm1 + a2 - a1, (idx1, i+k)

            if a3 - a2 > sm3 - sm2:
                sm3, idx3 = sm2 + a3 - a2, (*idx2, i+2*k)

        return idx3
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