<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]:
        n = len(nums)

        sums = [0] * n
        current_sum = sum(nums[:k])
        
        best_sums = [0 for i in range(n)]
        best_indexs = [[] for i in range(n)]
        for i in range(n-k):
            sums[i] = current_sum
            if i > 0 and best_sums[i-1] >= sums[i]:
                best_sums[i] = best_sums[i-1]
                best_indexs[i] = best_indexs[i-1]
            else:
                best_sums[i] = sums[i]
                best_indexs[i] = [i]
            current_sum = current_sum - nums[i] + nums[i + k]
        sums[n-k] = current_sum
        # print(sums, best_sums, best_indexs)

            
        for c in range(2):
            new_sums = [0 for i in range(n)]
            new_indexs = [[] for i in range(n)]
            for i in range(k, n-k+1):
                if new_sums[i-1] >= best_sums[i-k] + sums[i]:
                    new_sums[i] = new_sums[i-1]
                    new_indexs[i] = new_indexs[i-1]
                else:
                    new_sums[i] = best_sums[i-k] + sums[i]
                    new_indexs[i] = best_indexs[i-k][:] + [i]
            best_sums = new_sums
            best_indexs = new_indexs
            # print(best_sums, best_indexs)
        return best_indexs[n-k]


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