<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]:
        # Can use a K sized sliding window prefix sum to find the sum at each index
        # find the 3 start indexes whcih arnt overlapping that gives the max sum
        # Can use DFS for that

        window = 0
        psum = []
        total = 0
        for i, n in enumerate(nums):
            total += n
            window += 1
            if window >= k:
                psum.append(total)
                total -= nums[i-k+1]

        largest_left_of_i = {}
        largest_right_of_i = {}

        for i in range(0+k, len(psum)-k):
            j = i-k
            if largest_left_of_i and psum[j] <= psum[largest_left_of_i[i-1]]:
                largest_left_of_i[i] = largest_left_of_i[i-1]
            else:
                largest_left_of_i[i] = j

        for i in reversed(range(0+k, len(psum)-k)):
            j = i+k
            if largest_right_of_i and psum[j] < psum[largest_right_of_i[i+1]]:
                largest_right_of_i[i] = largest_right_of_i[i+1]
            else:
                largest_right_of_i[i] = j

        M = 0
        res = []

        for i in range(0+k, len(psum)-k):
            a = psum[largest_left_of_i[i]]
            b = psum[i]
            c = psum[largest_right_of_i[i]]
            if a + b + c > M:
                res = [largest_left_of_i[i], i, largest_right_of_i[i]]
                M = a + b + c

        return res



        # V = {}

        # def dfs(arr, i, x):
        #     # index arr left
        #     if (i,x) in V:
        #         return V[(i,x)]
        #     M = 0,[]
        #     for j in range(i, len(arr)):
        #         L = dfs(arr, j+k, x-1)
        #         if x == 0:
        #             curr = arr[j], [j]
        #         else:
        #             curr = arr[j] + L[0], [j] + L[1] 
        #         if curr[0] > M[0]:
        #             M = curr

        #     V[(i,x)] = M
        #     return M
        
        # return dfs(psum, 0, 2)[1]
        
        # want indexes
        # for all possible index after i return the greatest + running total
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