<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]:
        # Step 1: Compute the prefix sum
        n = len(nums)
        presum = [0] * (n + 1)
        for i in range(n):
            presum[i + 1] = presum[i] + nums[i]

        # Step 2: Prepare the DP table
        dp = [[(0, []) for _ in range(4)] for _ in range(n + 1)]

        # Step 3: Fill the DP table
        for i in range(1, n + 1):
            for j in range(1, 4):
                if i >= k:
                    # Case 1: Don't include the current subarray
                    sum_not_include_new = dp[i - 1][j][0]
                    
                    # Case 2: Include the current subarray
                    sum_include_new = dp[i - k][j - 1][0] + (presum[i] - presum[i - k])
                    
                    # Compare and decide which option is better
                    if sum_include_new > sum_not_include_new:
                        dp[i][j] = (sum_include_new, dp[i - k][j - 1][1] + [i - k])
                    else:
                        dp[i][j] = (sum_not_include_new, dp[i - 1][j][1])

        # The result is stored in dp[n][3][1] which has the list of starting indices for the three subarrays
        return dp[n][3][1]
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