<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:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();

        // build range sum
        rangeSum = vector<int>(n, 0);
        rangeSum[0] = accumulate(nums.begin(), nums.begin() + k, 0);

        for (int i = 1, j = k; j < n; ++i, ++j) {
            rangeSum[i] = rangeSum[i - 1] - nums[i - 1] + nums[j];
        }

        // build dp table for 1 subarray
        vector<vector<int>> dp1(n + 1);
        for (int i = n - k; 0 <= i; --i) {
            dp1[i] = {i};
            if (sum(dp1[i]) < sum(dp1[i + 1])) {
                dp1[i] = dp1[i + 1];
            }
        }

        // build dp table for 2 subarray
        vector<vector<int>> dp2(n + 1);
        for (int i = n - 2 * k; 0 <= i; --i) {
            dp2[i] = {i};
            dp2[i].insert(dp2[i].end(), dp1[i + k].begin(), dp1[i + k].end());
    
            if (sum(dp2[i]) < sum(dp2[i + 1])) {
                dp2[i] = dp2[i + 1];
            }
        }

        // build dp table for 3 subarray
        int res = n - 3;
        vector<vector<int>> dp3(n + 1);
        for (int i = n - 3 * k; 0 <= i; --i) {
            dp3[i] = {i};
            dp3[i].insert(dp3[i].end(), dp2[i + k].begin(), dp2[i + k].end());
    
            if (sum(dp3[i]) < sum(dp3[i + 1])) {
                dp3[i] = dp3[i + 1];
            } else {
                res = i;
            }
        }

        return dp3[res];
    }

private:
    vector<int> rangeSum;

    inline int sum(vector<int> &indice) {
        int res = 0;
        for (int idx : indice) {
            res += rangeSum[idx];
        }
        return res;
    }
};
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