<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]:
        new_nums = []
        total = 0
        for i in range(len(nums)):
            total += nums[i]
            
            if i >= k-1:
                new_nums.append(total)
                total -= nums[i-k+1]
        
        n = len(new_nums)
        
        left = []
        mx = -math.inf
        for i, num in enumerate(new_nums):
            if num > mx:
                mx = num
                idx = i
            left.append((mx, idx))
        
        right = [0] * n
        mx = -math.inf
        for i in range(len(new_nums) - 1, -1, -1):
            num = new_nums[i]
            if num >= mx:
                mx = num
                idx = i
            right[i] = (mx, idx)
        
        best = 0
        for i in range(n):
            left_idx = i - k
            right_idx = i + k
            if left_idx < 0 or right_idx >= n:
                continue
            
            total = left[left_idx][0] + new_nums[i] + right[right_idx][0]
            if total > best:
                best = total
                res = [left[left_idx][1], i, right[right_idx][1]]
        
        return res

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