<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]:
        biggestSumEndingUpToI = {} # maps i : (biggest total, leftmost starting index)
        
        curr = sum(nums[i] for i in range(k))
        biggestSum = curr
        biggestSumStartIndex = 0
        
        biggestSumEndingUpToI[k - 1] = (biggestSum, biggestSumStartIndex)
        
        for r in range(k, len(nums) - k):
            newNum = nums[r]
            lostNum = nums[r - k]
            curr += newNum
            curr -= lostNum
            if curr > biggestSum:
                biggestSum = curr
                biggestSumStartIndex = r - k + 1
            
            biggestSumEndingUpToI[r] = (biggestSum, biggestSumStartIndex)
        
        curr = sum(nums[i] for i in range(len(nums) - k, len(nums)))
        biggestSum = curr
        biggestSumStartIndex = len(nums) - k # leftmost
        
        biggestSumStartingAtI = {} # maps i : (biggest total, leftmost starting index) for [i:] sums
        
        biggestSumStartingAtI[len(nums) - k] = (biggestSum, biggestSumStartIndex)
        
        for l in range(len(nums) - k - 1, -1, -1):
            newNum = nums[l]
            lostNum = nums[l + k]
            curr += newNum
            curr -= lostNum
            if curr >= biggestSum:
                biggestSum = curr
                biggestSumStartIndex = l
            biggestSumStartingAtI[l] = (biggestSum, biggestSumStartIndex)        
        
        currSum = sum(nums[i] for i in range(k, 2 * k))
        leftBiggestSum, leftStartingIndex = biggestSumEndingUpToI[k - 1]
        rightBiggestSum, rightStartingIndex = biggestSumStartingAtI[2 * k]
        
        resBiggestSum = currSum + leftBiggestSum + rightBiggestSum
        res = [leftStartingIndex, k, rightStartingIndex]
        
        for midStart in range(k + 1, len(nums) - (2 * k) + 1):
            newRightEdge = midStart + k - 1
            newNum = nums[newRightEdge]
            lostNum = nums[midStart - 1]
            currSum += newNum
            currSum -= lostNum
            
            leftBiggestSum, leftStartingIndex = biggestSumEndingUpToI[midStart - 1]
            rightBiggestSum, rightStartingIndex = biggestSumStartingAtI[newRightEdge + 1]
            
            newTotalSum = currSum + leftBiggestSum + rightBiggestSum
            if newTotalSum > resBiggestSum:
                resBiggestSum = newTotalSum
                res = [leftStartingIndex, midStart, rightStartingIndex]
        
        return res
            
            
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