<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();
        nums.insert(nums.begin(), INT_MIN);
        vector<vector<int>> dp(n+1, vector<int>(4, INT_MIN/2));
        vector<vector<int>> track(n+1, vector<int>(4, -1));
        vector<int> presum(n+1);
        for (int i=1; i<n+1; i++)
            presum[i] = presum[i-1] + nums[i];
        for (int i=0; i<=n; i++)
        {
            dp[i][0] = 0;
            track[i][0] = i;
        }
        for (int j=1; j<=3; j++)
            for (int i=k; i<=n; i++)
            {
                int subsum = presum[i] - presum[i-k];
                if (dp[i-k][j-1] + subsum > dp[i-1][j])
                {
                    track[i][j] = i;
                    dp[i][j] = dp[i-k][j-1] + subsum;
                }
                else
                {
                    track[i][j] = track[i-1][j];
                    dp[i][j] = dp[i-1][j];
                }
            }
            vector<int> ret;
            int pre = track[n][3], j = 2;
            while (j >= 0 && pre >= 0)
            {
                ret.push_back(pre-k);
                pre = track[pre-k][j--];
            }
            reverse(ret.begin(), ret.end());
        return ret;
    }
};
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