<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)
        
        # Calculate prefix sums for efficient subarray sum computation
        prefix_sum = [0] * (n + 1)
        for i in range(n):
            prefix_sum[i + 1] = prefix_sum[i] + nums[i]
        
        # Function to get sum of subarray nums[i:i+k]
        def get_sum(i):
            return prefix_sum[i + k] - prefix_sum[i]
        
        # Initialize DP table
        dp = [[0] * 4 for _ in range(n + 1)]
        
        # Fill DP table
        for i in range(n - k + 1, -1, -1):
            for j in range(1, 4):
                if i + k <= n:
                    dp[i][j] = max(get_sum(i) + dp[i + k][j - 1], dp[i + 1][j])
                # else:
                #     dp[i][j] = dp[i + 1][j]

        # for i in range(n):
        #     for j in range(1, 4):
        #         if i + k <= n:
        #             dp[i][j] = max(dp[i][j], dp[i - k][j - 1] + get_sum(i))
        #         else:
        #             dp[i][j] = dp[i - 1][j]
        
        # Backtrack to find the starting indices
        ans = []
        i, j = 0, 3
        while j > 0:
            if i + k <= n and dp[i][j] == get_sum(i) + dp[i + k][j - 1]:
                ans.append(i)
                i += k
                j -= 1
            else:
                i += 1
        
        return ans
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