<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) 
        psa = [0] + list(accumulate(nums))
        nums = [0] + nums
        dp = [[0] * (3 + 1) for _ in range(n + 1)]
         
        for i in range(k, n + 1):
            for j in range(1, 3 + 1):  
                dp[i][j] = max(dp[i - 1][j], dp[i - k][j - 1] + psa[i] - psa[i - k])
         
        i = n; j = 3; ans = []
        while i >= k and j > 0: # 字典序最小
            if dp[i - 1][j] == dp[i][j]: i -= 1 # 能不选，则不选
            else:
                ans.append(i - k)
                i -= k
                j -= 1
        return ans[::-1]
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