<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, m=3) -> List[int]:

        presum = [0]
        for i in nums:
            presum.append(presum[-1]+i)
        
        n = len(nums)
        dp = [[0]*(n+1) for i in range(m+1)]
        index = [[0]*(n+1) for i in range(m+1)]

        for i in range(1, m+1):
            for j in range(i*k, n+1):
                cur = dp[i-1][j-k]+presum[j]-presum[j-k]
                if cur>dp[i][j-1]:
                    dp[i][j] = cur
                    index[i][j] = j-k
                else:
                    dp[i][j]=dp[i][j-1]
                    index[i][j]=index[i][j-1]
        
        res = [0]*m
        res[-1] = index[m][n]
        for i in range(m-2, -1, -1):
            res[i] = index[i+1][res[i+1]]
        return res
        
        



        
        
            
        

        


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