<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)
        prefix = [0]*(n+1)
        for i in range(1, n+1):
            prefix[i] = prefix[i-1] + nums[i-1]
        
        dp = [[0 for _ in range(n+1)] for j in range(4)]
        idx = [[0 for _ in range(n+1)] for j in range(4)]
        m = 3

        # use dp[0][0] as base case to build from
        for i in range(1, m+1):
            for j in range(i*k, n-(m-i)*k+1):
                # if we consider skip current number
                dp[i][j] = dp[i][j-1]
                # previous starting point would also be current starting point
                idx[i][j] = idx[i][j-1]
                # if we decide to take current number instead
                # then it is previous best results (minus k) add current sum
                curr = prefix[j] - prefix[j-k] + dp[i-1][j-k]
                if curr > dp[i][j]:
                    dp[i][j] = curr
                    idx[i][j] = j-k
        
        res = [0]*m
        last = n

        # for last index, it must be the last element of idx last row, let's say it's index 5, which is the 6th element
        # which means the m-1 subarray ends before index 5 (or ends with at most element 5)
        # so dp[m-1][5] gets corresponding value and idx[m-1][5] gets starting index
        for i in range(m, 0, -1):
            res[i-1] = idx[i][last]
            last = res[i-1]
        
        return res


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