<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]:
        
        
        
        # my solution ... 212 ms ... 81 % ... 17.4 MB ... 78 %
        #  time: O(n)
        # space: O(n) ... 可优化至 O(k)
        
        n = len(A)
        winsum = sum(A[:k])
        maxone, maxtwo, maxthree = [0]*n, [0]*n, [0]*n    # maxone[i] = (sidx, wsum) 表示 A[:i+1] 中最大的 wsum 及相应起始索引 sidx
        maxone[k-1] = (0, winsum)                         # maxtwo[i] = (idx1, idx2, wsum)    maxthree[i] = (idx1, idx2, idx3, wsum)
        for i in range(k, n):
            winsum += A[i] - A[i-k]                       # 更新 winsum
            
            if maxone[i-1][1] < winsum:                   # 更新 maxone
                maxone[i] = (i-k+1, winsum)
            else:
                maxone[i] = maxone[i-1]
            
            if i == 2*k-1:                                # 更新 maxtwo
                maxtwo[i] = (0, k, maxone[i-k][1] + winsum)
            elif i > 2*k-1:
                if maxtwo[i-1][2] < maxone[i-k][1] + winsum:
                    maxtwo[i] = (maxone[i-k][0], i-k+1, maxone[i-k][1] + winsum)
                else:
                    maxtwo[i] = maxtwo[i-1]
            
            if i == 3*k-1:                                # 更新 maxthree
                maxthree[i] = (0, k, 2*k, maxtwo[i-k][2] + winsum)
            elif i > 3*k-1:
                if maxthree[i-1][3] < maxtwo[i-k][2] + winsum:
                    maxthree[i] = (maxtwo[i-k][0], maxtwo[i-k][1], i-k+1, maxtwo[i-k][2] + winsum)
                else:
                    maxthree[i] = maxthree[i-1]
        
        return list(maxthree[-1])[:-1]
        
        
        
        
        
        
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