<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]:
        ksum = []
        cur = 0
        for i, v in enumerate(nums):
            cur += v
            if i>=k:
                cur -= nums[i-k]
            if i>=k-1:
                ksum.append(cur)
        L = len(ksum)
        left = [0] * L
        for i in range(1,L):
            if ksum[i] > ksum[left[i-1]]:
                left[i] = i
            else:
                left[i] = left[i-1]
        right = [L-1] *L
        for i in reversed(range(L-1)):
            if ksum[i] >= ksum[right[i+1]]:
                right[i] =i
            else:
                right[i] = right[i+1]

        res = None
        max3 = -math.inf
        for j in range(k,L-k):
            i = left[j-k]
            m = right[j+k]
            val = ksum[i] + ksum[j] + ksum[m]
            if val > max3:
                res = [i,j,m]
                max3 = val
        print(ksum)
        print(left)
        print(right)
        return res


[3, 3, 3, 8, 13, 12, 6]
[0, 0, 0, 3, 4, 4, 4]
[4, 4, 4, 4, 4, 5, 6]


[3, 3, 3, 8, 13, 12, 6]
[0, 0, 0, 3, 4, 4, 4]
[4, 4, 4, 4, 4, 5, 6]
20400