<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]:
        prefix_sum=[0]
        for x in nums:
            prefix_sum.append(prefix_sum[-1]+x)
        n=len(prefix_sum)
        best=[]
        max_sum=0
        dp=[[0]*n for _ in range(4)]
        def maxSums(i, curSum,arr,x):
            nonlocal max_sum
            nonlocal best
            if dp[x][i]>curSum:
                return 
            dp[x][i]=curSum
            if i+k>n-1 or x==0:
                if curSum>=max_sum:
                    max_sum=curSum
                    best=arr.copy()
                return curSum
            x1=maxSums(i+1, curSum, arr,x)
            arr.append(i)
            x2=maxSums(i+k, curSum+prefix_sum[i+k]-prefix_sum[i], arr,x-1)
            arr.pop()
        maxSums(0, 0, [],3)
        return best
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