<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();
        vector<vector<int>> dp(n, vector<int>(3, 0));
        vector<vector<int>> pos(n, vector<int>(3, -1));
        vector<int> ans(3, -1);
        for (int i = 0, kSum = 0; i < n; i++) {
            kSum += nums[i];
            if (i >= k)
                kSum -= nums[i - k];
            if (i >= k -1) {
                dp[i][0]  = i > 0 ? dp[i - 1][0] : 0;
                pos[i][0] = i > 0 ? pos[i - 1][0] : -1;
                if (dp[i][0] < kSum)
                    dp[i][0] = kSum, pos[i][0] = i;
            }
            if (i >= 2 * k - 1) {
                dp[i][1]  = dp[i - 1][1];
                pos[i][1] = pos[i - 1][1];
                if (int tmp = kSum + dp[i - k][0]; tmp > dp[i][1])
                    dp[i][1] = tmp, pos[i][1] = i;
            }
            if (i >= 3 * k - 1) {
                dp[i][2]  = dp[i - 1][2];
                pos[i][2] = pos[i - 1][2];
                if (int tmp = kSum + dp[i - k][1]; tmp > dp[i][2])
                    dp[i][2] = tmp, pos[i][2] = i;
            }
        }
        // for (int i = 0; i < n; i++) cout << dp[i][0] << ", "; cout << endl;
        // for (int i = 0; i < n; i++) cout << dp[i][1] << ", "; cout << endl;
        // for (int i = 0; i < n; i++) cout << dp[i][2] << ", "; cout << endl;

        // for (int i = 0; i < n; i++) cout << pos[i][0] << ", "; cout << endl;
        // for (int i = 0; i < n; i++) cout << pos[i][1] << ", "; cout << endl;
        // for (int i = 0; i < n; i++) cout << pos[i][2] << ", "; cout << endl;

        for (int j = 2, end = pos[n - 1][2]; j >= 0; j--) {
            ans[j] = end - k + 1;  // start pos of current subarray
            if (j > 0) end = pos[end - k][j - 1];
        }
        return ans;
    }
};
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