<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 {
public:
    std::vector<int> maxSumOfThreeSubarrays(const std::vector<int>& nums, int k) {
        int n = nums.size();
        std::vector<int> partialSum(n + 1, 0);
        for (int i = 0; i < n; ++i)
            partialSum[i + 1] = partialSum[i] + nums[i];
        // Now, sum(nums[i]~nums[j]) = partialSum[j + 1] - partialSum[i].
        auto sum = [&](int i, int j) -> int {
            return partialSum[j + 1] - partialSum[i];
        };
        // Now, let's use dynamic programming to solve the problem.
        std::array<std::vector<int>, 3> maxSums = {{
            std::vector<int>(n + 1, 0),
            std::vector<int>(n + 1, 0),
            std::vector<int>(n + 1, 0)
        }};
        // maxSums[t][i] means the max sum of t+1 subarrays in nums[i]~nums.back()
        std::array<std::vector<std::array<int, 3>>, 3> maxSumSolutions = {{
            std::vector<std::array<int, 3>>(n + 1),
            std::vector<std::array<int, 3>>(n + 1),
            std::vector<std::array<int, 3>>(n + 1)
        }};
        // maxSumSolutions[t][i] means the solution for maxSums[t][i].
        for (int i = n - k; i >= 0; --i) {
            if (sum(i, i + k - 1) + maxSums[1][i + k] >= maxSums[2][i + 1]) {
                maxSums[2][i] = sum(i, i + k - 1) + maxSums[1][i + k];
                maxSumSolutions[2][i] = {{ i, maxSumSolutions[1][i + k][0], maxSumSolutions[1][i + k][1] }};
            } else {
                maxSums[2][i] = maxSums[2][i + 1];
                maxSumSolutions[2][i] = maxSumSolutions[2][i + 1];
            }
            if (sum(i, i + k - 1) + maxSums[0][i + k] >= maxSums[1][i + 1]) {
                maxSums[1][i] = sum(i, i + k - 1) + maxSums[0][i + k];
                maxSumSolutions[1][i] = {{ i, maxSumSolutions[0][i + k][0], -1 }};
            } else {
                maxSums[1][i] = maxSums[1][i + 1];
                maxSumSolutions[1][i] = maxSumSolutions[1][i + 1];
            }
            if (sum(i, i + k - 1)                     >= maxSums[0][i + 1]) {
                maxSums[0][i] = sum(i, i + k - 1);
                maxSumSolutions[0][i] = {{ i, -1, -1 }};
            } else {
                maxSums[0][i] = maxSums[0][i + 1];
                maxSumSolutions[0][i] = maxSumSolutions[0][i + 1];
            }
        }
        // Get the answer.
        return { maxSumSolutions[2][0][0], maxSumSolutions[2][0][1], maxSumSolutions[2][0][2] };
    }
};
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