<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<long long>preNums(n, 0);
        preNums[0] = nums[0];
        for(int i=1;i<n;i++)
            preNums[i] = preNums[i-1]+nums[i];

        vector<pair<long long,int>>prevBest(n, {0,0});
        prevBest[k-1] = {preNums[k-1], 0};
        for(int i=k;i<n;i++) {
            long long x = preNums[i]-preNums[i-k];
            if(x>prevBest[i-1].first)
                prevBest[i] = {x, i-k+1};
            else
                prevBest[i] = prevBest[i-1];
        }

        vector<pair<long long,int>>nextBest(n, {0,0});
        nextBest[n-k] = {preNums[n-1]-preNums[n-k-1], n-k};
        for(int i=n-k-1;i>=0;i--) {
            long long x = preNums[i+k-1]-(i==0?0:preNums[i-1]);
            if(x>=nextBest[i+1].first)
                nextBest[i] = {x, i};
            else
                nextBest[i] = nextBest[i+1];
        }

        vector<int>res(3, -1);
        long long maxSum = 0;
        for(int i=k;i<=n-2*k;i++) {
            long long x = prevBest[i-1].first + preNums[i+k-1]-preNums[i-1] + nextBest[i+k].first;
            if(x>maxSum) {
                maxSum = x;
                res = {prevBest[i-1].second, i, nextBest[i+k].second};
            }
        }
        return res;
    }
};
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