<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>

from typing import List

class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        m, n = len(nums), 3
        dp = [[[0, -1] for _ in range(m)] for _ in range(n + 1)]
        prefix_sum = [0] * m
        curr_sum = 0

        for index, num in enumerate(nums):
            curr_sum += num
            prefix_sum[index] = curr_sum

        for i in range(1, n + 1):
            for j in range(k - 1, m):
                ai = prefix_sum[j] - prefix_sum[j - k] + dp[i - 1][j - k][0] if j - k >= 0 else prefix_sum[j]
                if ai > dp[i][j - 1][0]:
                    dp[i][j] = [ai, j - k + 1]
                else:
                    dp[i][j] = [dp[i][j-1][0], dp[i][j - 1][1]]

        third = dp[n][m-1][1]
        second = dp[n-1][third-1][1]
        first = dp[n-2][second-1][1]
        return [first, second, third]
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