<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<int>dp(n+1,0),suff(n+1,0);
        for(int i=n-1;i>=0;i--)suff[i]=suff[i+1]+nums[i];
        for(int i=n-k;i>=0;i--)dp[i]=suff[i]-suff[i+k];
        set<pair<int,int>>bef,aft;
        bef.insert({dp[0],0});
        for(int i=2*k;i<=n-k;i++)aft.insert({dp[i],-i});
        vector<int>ans(3);
        int sum = 0;
        for(int i=k;i<=n-2*k;i++){
            auto l = *bef.rbegin();
            auto r = *aft.rbegin();
            int cur = dp[i]+l.first+r.first;
            if(cur>sum){
                sum = cur;
                ans[0]=-l.second;
                ans[1]=i;
                ans[2]=-r.second;
            }
            bef.insert({dp[i-k+1],-(i-k+1)});
            aft.erase({dp[i+k],-(i+k)});
        }
        cout<<sum<<endl;
        return ans;

    }
};
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