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
func maxSumOfThreeSubarrays(nums []int, k int) []int {
	ans := make([]int, 3)
	s, s1, s2, s3 := 0, 0, 0, 0
	mx1, mx12 := 0, 0
	idx1, idx121, idx122 := 0, 0, 0
	for i := k * 2; i < len(nums); i++ {
		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
				idx121 = idx1
				idx122 = i - k*2 + 1
			}
			if mx12+s3 > s {
				s = mx12 + s3
				ans = []int{idx121, idx122, i - k + 1}
			}
			s1 -= nums[i-k*3+1]
			s2 -= nums[i-k*2+1]
			s3 -= nums[i-k+1]
		}
	}
	return ans
}