<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) {
        
        if (nums.empty() || k == 0) return {};

        int N = nums.size();
        int rec = 0;
        int window = 0;
        int maxSum = 0;
        int maxHead = 0;
        int headpos = 0;
        vector<int> ans(3, 0);
        vector<pair<int, int>> redp(N, {0, 0});

        redp[N - k].first = N - k;
        for (int i = N - 1; i >= N - k; i--) {
            redp[N - k].second += nums[i];
        }

        rec = redp[N - k].second;
        for (int i = N - k - 1; i >= (k << 1); i--) {
            rec = rec + nums[i] - nums[i + k];
            if (rec >= redp[i + 1].second) {
                redp[i] = {i, rec};
            } else {
                redp[i] = redp[i + 1];
            }
        }

        rec = 0;
        for (int i = 0; i < k; i++) {
            rec += nums[i];
            window += nums[i + k];
        }

        ans = {0, k, redp[(k << 1)].first};
        maxSum = rec + window + redp[(k << 1)].second;
        maxHead = rec;
        for (int i = k + 1; i <= (N - (k << 1)); i++) {
            rec = rec + nums[i - 1] - nums[i - 1 - k];
            if (rec > maxHead) {
                maxHead = rec;
                headpos = i - k;
            }
            window = window + nums[i + k - 1] - nums[i - 1];
            int curans = maxHead + window + redp[i + k].second;
            if (curans > maxSum) {
                maxSum = curans;
                ans = {headpos, i, redp[i + k].first};
            }
        }

        return ans;
    }
};
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