<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:
    /**
     * [1,2,1,2,6,7,5,1]
     * [3,3,3,8,13,12,6]
     */
    vector<int> GetSumWithLengthK(vector<int>& nums, int k) {
        int left = 0;
        vector<int> res;
        int sum = 0;
        for (int i = 0; i < nums.size(); ++i) {
            sum += nums[i];
            if (i - left == k) {
                sum -= nums[left];
                ++left;
            }
            if (i - left + 1 == k) {
                res.push_back(sum);
            }
        }
        return res;
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        vector<int> vecSum = GetSumWithLengthK(nums, k);

        int n = vecSum.size();
        vector<vector<pair<int, vector<int>>>> dp(n, vector<pair<int, vector<int>>>(4));
        dp[0][1] = {vecSum[0], {0}};
        for (int idx = 1; idx < n; idx++) {
            if (vecSum[idx] > dp[idx - 1][1].first) {
                dp[idx][1] = {vecSum[idx], {idx}};
            } else {
                dp[idx][1] = dp[idx - 1][1];
            }
            for (int i = 2; i <= 3; ++i) {
                // dp[idx][i] = max(dp[idx - 1][i], dp[idx - k][i - 1] + vecSum[idx]);
                if (idx >= k && (dp[idx - k][i - 1].first + vecSum[idx] > dp[idx - 1][i].first)) {
                    dp[idx][i] = dp[idx - k][i - 1];
                    dp[idx][i].first += vecSum[idx];
                    dp[idx][i].second.push_back(idx);
                } else {
                    dp [idx][i] = dp[idx - 1][i];
                }
            }
        }

        return dp[n - 1][3].second;
    }
};
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