<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 makeS(self,nums,k):
        s = sum(nums[:k])
        S = [ s ]
        for i in range(k,len(nums)):
            s += nums[i] - nums[i-k]
            S.append( s )
        return S
    def maxSumOfThreeSubarrays(self, nums, k, m = 3):
        L = len(nums)
        if (k*m)>=L:
            return sum( nums )
        #
        S    = self.makeS(nums,k)
        Aprev = [ [0,[]] for _ in range(L) ]
        for i in range(m):
            Anew  = [ [ Aprev[0][0]+S[i*k], Aprev[0][1] + [i*k] ] ]
            for j in range(i*k+1,len(S)):
                sp,ip = Aprev[j-i*k]
                s = S[j] + sp
                if s>Anew[-1][0]:
                    Anew.append( [s, ip + [j] ] )
                else:
                    Anew.append( Anew[-1] )
            Aprev = Anew
        return Anew[-1][-1]
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