Maximum Sum of 3 Non-Overlapping Subarrays
<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>
Array; Dynamic Programming; Prefix Sum; Sliding Window
class Solution:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        s = s1 = s2 = s3 = 0
        mx1 = mx12 = 0
        idx1, idx12 = 0, ()
        ans = []
        for i in range(k * 2, len(nums)):
            s1 += nums[i - k * 2]
            s2 += nums[i - k]
            s3 += nums[i]
            if i >= k * 3 - 1:
                if s1 > mx1:
                    mx1 = s1
                    idx1 = i - k * 3 + 1
                if mx1 + s2 > mx12:
                    mx12 = mx1 + s2
                    idx12 = (idx1, i - k * 2 + 1)
                if mx12 + s3 > s:
                    s = mx12 + s3
                    ans = [*idx12, i - k + 1]
                s1 -= nums[i - k * 3 + 1]
                s2 -= nums[i - k * 2 + 1]
                s3 -= nums[i - k + 1]
        return ans
