<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]:
        n = len(nums)
        
        # Prefix sum array
        prefix_sum = [0] * (n + 1)
        for i in range(n):
            prefix_sum[i + 1] = prefix_sum[i] + nums[i]
        
        # Best single subarray sum and its ending index
        max_sum = [0] * (n + 1)
        max_sum_index = [-1] * (n + 1)
        for i in range(k, n + 1):
            current_sum = prefix_sum[i] - prefix_sum[i - k]
            if current_sum > max_sum[i - 1]:
                max_sum[i] = current_sum
                max_sum_index[i] = i - k
            else:
                max_sum[i] = max_sum[i - 1]
                max_sum_index[i] = max_sum_index[i - 1]
        
        # Best two subarrays sum and their ending indices
        max_two_sums = [0] * (n + 1)
        max_two_sums_indices = [(-1, -1)] * (n + 1)
        for i in range(2 * k, n + 1):
            current_sum = prefix_sum[i] - prefix_sum[i - k]
            best_previous = max_sum[i - k]
            best_previous_index = max_sum_index[i - k]
            if current_sum + best_previous > max_two_sums[i - 1]:
                max_two_sums[i] = current_sum + best_previous
                max_two_sums_indices[i] = (best_previous_index, i - k)
            else:
                max_two_sums[i] = max_two_sums[i - 1]
                max_two_sums_indices[i] = max_two_sums_indices[i - 1]
        
        # Best three subarrays sum and their ending indices
        max_three_sums = [0] * (n + 1)
        max_three_sums_indices = [(-1, -1, -1)] * (n + 1)
        for i in range(3 * k, n + 1):
            current_sum = prefix_sum[i] - prefix_sum[i - k]
            best_previous = max_two_sums[i - k]
            best_previous_indices = max_two_sums_indices[i - k]
            if current_sum + best_previous > max_three_sums[i - 1]:
                max_three_sums[i] = current_sum + best_previous
                max_three_sums_indices[i] = (*best_previous_indices, i - k)
            else:
                max_three_sums[i] = max_three_sums[i - 1]
                max_three_sums_indices[i] = max_three_sums_indices[i - 1]
        
        # The result is the last entry in max_three_sums_indices
        return list(max_three_sums_indices[n])

# Example usage:
solution = Solution()
nums = [1, 2, 1, 2, 6, 7, 5, 1]
k = 2
print(solution.maxSumOfThreeSubarrays(nums, k))  # Output: [0, 3, 5]
24100