<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, a: List[int], k: int) -> List[int]:
        MAX = 3
        n = len(a)
        dp = [[[-1,-1] for _ in range(n)] for i in range(MAX)]
        pre = [0]
        for i in a:
            pre.append(pre[-1]+i)
        def fun(i):
            if i+k<=n:
                return pre[i+k]-pre[i]
            return -inf
        for i in range(MAX):
            start = i*k
            for j in range(i*k,n):
                if i==0:
                    val = fun(j)
                else:
                    val = fun(j)+dp[i-1][j-k][1]
                if j==start or val>dp[i][j-1][1]:
                    dp[i][j] = [j,val]
                else:
                    dp[i][j] = dp[i][j-1][:]
        ind = n-1
        ans = []
        for i in range(MAX-1,-1,-1):
            ind = dp[i][ind][0]
            ans.append(ind)
            ind = ind-k
        return ans[::-1]


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