<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]:
        self.array=[0]*len(nums)
        self.array[0]=sum(nums[:k])
        for i in range(1,len(nums)):
            if k+i-1<len(nums):self.array[i]=self.array[i-1]+nums[k+i-1]-nums[i-1]

       
        self.dp=[[0]*(len(nums)+1) for _ in range(4)]
        for i in range(1,4):
            for j in range(1,len(nums)+1):
                if j<i*k: self.dp[i][j]=self.dp[i-1][j]
                else:
                   self.dp[i][j]=max(self.dp[i][j-1],self.dp[i-1][j-k]+self.array[j-k])
        
        i=3
        j=len(nums)
        res=[0]*3
        while(j>0):
            if self.dp[i][j]==self.dp[i][j-1]: j-=1
            else:
                j-=k
                i-=1
                res[i]=j
            if i==0 or j==0: break
        return res
                


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