<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):
        n, w1, w2, w3 = len(nums), sum(nums[:k]), sum(nums[k:2*k]), sum(nums[2*k:3*k])
        total1, total2, total3 = w1, w1+w2, w1+w2+w3
        total1_idx, total2_idx, total3_idx = [0], [0,k], [0,k,2*k]

        for i in range(1,n-3*k+1):
            w1 += nums[i-1+k] - nums[i-1]
            w2 += nums[i-1+2*k] - nums[i-1+k]
            w3 += nums[i-1+3*k] - nums[i-1+2*k]

            if w1 > total1:
                total1, total1_idx = w1, [i]

            if total1 + w2 > total2:
                total2, total2_idx = total1 + w2, total1_idx + [i+k]

            if total2 + w3 > total3:
                total3, total3_idx = total2 + w3, total2_idx + [i+2*k]

        return total3_idx

        
        
        
19800