<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]:
        # https://www.youtube.com/watch?v=oo_T4GdRot4
        presum = []
        for i, num in enumerate(nums):
            p = num + (presum[-1] if i > 0 else 0)
            presum.append(p)

        n = len(nums)

        leftMax, leftIdx = [0]*n, [0]*n
        maxSum, maxIdx = 0, 0
        for i in range(k-1, n):
            cursum = presum[i] - (presum[i-k] if i >= k else 0)
            if cursum > maxSum:
                maxSum = cursum
                maxIdx = i-k+1
            leftMax[i] = maxSum
            leftIdx[i] = maxIdx
        
        rightMax, rightIdx = [0]*n, [0]*n
        maxSum, maxIdx = 0, 0
        for i in range(n-k, -1, -1):
            cursum = presum[i+k-1] - (presum[i-1] if i!=0 else 0)
            if cursum >= maxSum:
                maxSum = cursum
                maxIdx = i
            rightMax[i] = maxSum
            rightIdx[i] = maxIdx

        print(leftMax)
        print(rightMax)
        res = []
        maxSum = 0
        for i in range(k, n-2*k+1):
            mid = presum[i+k-1]-(presum[i-1] if i!=0 else 0)
            cursum = leftMax[i-1] + mid + rightMax[i+k]
            if cursum > maxSum:
                maxSum = cursum
                res = [leftIdx[i-1], i, rightIdx[i+k]]
        return res

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