<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)
        
        # Precompute sums of subarrays of length k
        prefix_sums = [0] * (n - k + 1)
        window_sum = sum(nums[:k])
        prefix_sums[0] = window_sum
        for i in range(1, n - k + 1):
            window_sum = window_sum - nums[i-1] + nums[i+k-1]
            prefix_sums[i] = window_sum
        
        # Memoization dictionary
        memo = {}
        
        def dp(i, m):
            if m == 0:
                return 0, []
            if i > n - m*k:
                return float('-inf'), []
            
            if (i, m) in memo:
                return memo[(i, m)]
            
            # Don't include current subarray
            dont_take_sum, dont_take_indices = dp(i + 1, m)
            
            # Include current subarray
            take_sum, take_indices = dp(i + k, m - 1)
            take_sum += prefix_sums[i]
            take_indices = [i] + take_indices
            
            if take_sum > dont_take_sum or (take_sum == dont_take_sum and take_indices < dont_take_indices):
                result = (take_sum, take_indices)
            else:
                result = (dont_take_sum, dont_take_indices)
            
            memo[(i, m)] = result
            return result
        
        max_sum, indices = dp(0, 3)
        return indices
35600