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
function maxSumOfThreeSubarrays(nums: number[], k: number): number[] {
    const n: number = nums.length;
    const s: number[] = Array(n + 1).fill(0);

    for (let i = 0; i < n; ++i) {
        s[i + 1] = s[i] + nums[i];
    }

    const pre: number[][] = Array(n)
        .fill([])
        .map(() => new Array(2).fill(0));
    const suf: number[][] = Array(n)
        .fill([])
        .map(() => new Array(2).fill(0));

    for (let i = 0, t = 0, idx = 0; i < n - k + 1; ++i) {
        const cur: number = s[i + k] - s[i];
        if (cur > t) {
            pre[i + k - 1] = [cur, i];
            t = cur;
            idx = i;
        } else {
            pre[i + k - 1] = [t, idx];
        }
    }

    for (let i = n - k, t = 0, idx = 0; i >= 0; --i) {
        const cur: number = s[i + k] - s[i];
        if (cur >= t) {
            suf[i] = [cur, i];
            t = cur;
            idx = i;
        } else {
            suf[i] = [t, idx];
        }
    }

    let ans: number[] = [];
    for (let i = k, t = 0; i < n - 2 * k + 1; ++i) {
        const cur: number = s[i + k] - s[i] + pre[i - 1][0] + suf[i + k][0];
        if (cur > t) {
            ans = [pre[i - 1][1], i, suf[i + k][1]];
            t = cur;
        }
    }

    return ans;
}
