<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]:
        subSums = list()
        subSum = 0
        for i, num in enumerate(nums):
            subSum += num
            if i >= k:
                subSum -= nums[i - k]
            subSums.append(subSum)
        subMax = [[(0,0)]*len(nums) for _ in range(4)]
        
        print(nums)
        print(subSums)
        print()
        for i in range(1, 4):
            current = 0
            lastIdx = 0
            for j in range(k*i - 1, len(nums)):
                nextSum = subMax[i-1][j-k][0] + subSums[j]
                if nextSum > current:
                    lastIdx = j
                    current = nextSum
                subMax[i][j] = (current, lastIdx)
            print(subMax[i])

        result = list()
        idx = len(nums)-1
        for i in range(3, 0, -1):
            right = subMax[i][idx][1]
            result.append(right - k + 1)
            idx = right - k
        return result[::-1]
25700