<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]:        
        # get all k-window sums of nums
        ksums = []
        window = sum(nums[:k])
        ksums.append(window)
        for i in range(k, len(nums)):
            window += nums[i]
            window -= nums[i - k]
            ksums.append(window)
        
        # editorial: for each j, find i->[0, j-k] and l->[j+k, N]
        # where i/l are indices of (first) max value in left/right range
        # i.e. ksums[i] + ksums[l] is maximized
        # answer is whichever (i, j, l) gives max sum and lexicographically least
        
        rightMaxes = dict()  # l => first idx of max ksum in ksums[l:]
        rightMaxIdx = len(ksums) - 1
        for l in range(len(ksums) - 1, 2*k - 1, -1):
            if ksums[l] >= ksums[rightMaxIdx]:
                rightMaxIdx = l
            rightMaxes[l] = rightMaxIdx
        
        print(rightMaxes)
        
        triplets = []
        leftMaxIdx = 0
        for j in range(k, len(ksums) - k):
            curleft = j - k
            curright = j + k
            
            if ksums[curleft] > ksums[leftMaxIdx]:
                leftMaxIdx = curleft
            rightMaxIdx = rightMaxes[curright]
            
            triplets.append((leftMaxIdx, j, rightMaxIdx))
        
        maxSum = 0
        for i, j, l in triplets:
            maxSum = max(maxSum, ksums[i] + ksums[j] + ksums[l])
        
        for i, j, l in triplets:
            if ksums[i] + ksums[j] + ksums[l] == maxSum:
                return [i, j, l]
        assert False
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