<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]:
        # MISTAKE: 1-tuples require comma, e.g.: (val,)
        curr = sum(nums[:k])

        max_1_sum = [0] * len(nums)
        max_2_sum = [0] * len(nums)
        max_3_sum = [0] * len(nums)

        max_1_ind = [(0,)] * len(nums)
        max_2_ind = [(0, k)] * len(nums)
        max_3_ind = [(0, k, 2 * k)] * len(nums)

        max_1_sum[k - 1] = curr

        for i in range(k, len(nums)):
            curr += nums[i] - nums[i - k]
            if curr > max_1_sum[i - 1]:
                max_1_ind[i] = (i - k + 1,)
                max_1_sum[i] = curr
            else:
                max_1_ind[i] = max_1_ind[i - 1]
                max_1_sum[i] = max_1_sum[i - 1]

            if i >= 2 * k - 1:
                if max_1_sum[i - k] + curr > max_2_sum[i - 1]:
                    max_2_ind[i] = (max_1_ind[i - k][0], i - k + 1)
                    max_2_sum[i] = max_1_sum[i - k] + curr
                else:
                    max_2_ind[i] = max_2_ind[i - 1] 
                    max_2_sum[i] = max_2_sum[i - 1]

            if i >= 3 * k - 1:
                if max_2_sum[i - k] + curr > max_3_sum[i - 1]:
                    max_3_ind[i] = (max_2_ind[i - k][0], max_2_ind[i - k][1], i - k + 1)
                    max_3_sum[i] = max_2_sum[i - k] + curr
                else:
                    max_3_ind[i] = max_3_ind[i - 1] 
                    max_3_sum[i] = max_3_sum[i - 1]
            
        return list(max_3_ind[-1])
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