<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 in range(len(nums)):
            cur += nums[i]
            if i >=k:
                cur -= nums[i-k]
            if i >=k-1:
                ksum.append(cur)
        L = len(ksum)  # len(nums) - (k-1)
        maxSum = [[0] * L for _ in range(4)] 
        startPos = [[0] * L for _ in range(4)]
        # maxSum[0] = ksum
        # startPos[0] = list(range(L))

        for i in range(1,3+1):
            for j in range(L):
                if i == 1 and j == 0:
                    maxSum[i][j] = ksum[j]
                elif i == 1 and j > 0:
                    if ksum[j] > maxSum[i][j-1]:
                        startPos[i][j] = j
                        maxSum[i][j] = ksum[j]
                    else:
                        startPos[i][j] =  startPos[i][j-1]
                        maxSum[i][j] = maxSum[i][j-1]
                elif i > 1:
                    if j>=k and ksum[j] + maxSum[i-1][j-k] > maxSum[i][j-1]:
                        startPos[i][j] = j
                        maxSum[i][j] = ksum[j] + maxSum[i-1][j-k]
                    else:
                        startPos[i][j] =  startPos[i][j-1]
                        maxSum[i][j] = maxSum[i][j-1]
        res3 = startPos[3][L-1]
        res2 = startPos[2][res3-k]
        res1 = startPos[1][res2-k]
        print(ksum)
        print(maxSum)
        print(startPos)
        return [res1, res2, res3]

nums = \
[7,13,20,19,19,2,10,1,1,19]

k = 3

ksum = \
[40, 52, 58, 40, 31, 13, 12, 21]

maxSum = \
[[0, 0, 0, 0, 0, 0, 0, 0], 
 [40, 52, 58, 58, 58, 58, 58, 58], 
 [0, 0, 0, 80, 83, 83, 83, 83], 
 [0, 0, 0, 40, 40, 40, 92, 104]]

startPos = \
[[0, 0, 0, 0, 0, 0, 0, 0], 
 [0, 1, 2, 2, 2, 2, 2, 2], 
 [0, 0, 0, 3, 4, 4, 4, 4], 
 [0, 0, 0, 3, 3, 3, 6, 7]]

       
23500