<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:

    int psum[200000];

    void build(vector<int> &nums) {
        psum[0] = nums[0];
        for(int i=1;i<nums.size();i++) psum[i] = psum[i-1] + nums[i];
    }

    int query(int i, int j) {
        int sum = psum[j];
        if(i > 0) sum -= psum[i-1];
        return sum;
    }

    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        build(nums);
        vector<int> sum1(n, -1);
        for(int i=n-k;i>=0;i--) sum1[i] = query(i, i + k - 1);
        map<int, int> mp;
        vector<pair<int, int>> sum2(n, {-1, -1});
        int ind = n - k;
        for(int i=n-2*k;i>=0;i--) {
            mp[sum1[ind]] = ind;
            auto key = mp.rbegin();
            pair<int, int> p = *key;
            sum2[i].first = query(i, i + k - 1) + p.first;
            sum2[i].second = p.second;
            ind--;
        }
        vector<pair<int, int>> sum3(n, {-1, -1});
        mp.clear();
        ind = n - 2 * k;
        for(int i=n-3*k;i>=0;i--) {
            mp[sum2[ind].first] = ind;
            auto key = mp.rbegin();
            pair<int, int> p = *key;
            sum3[i].first = query(i, i + k - 1) + p.first;
            sum3[i].second = p.second;
            ind--;
        }
        int ans = 0;
        vector<int> p = {-1, -1, -1};
        for(int i=0;i<=n-3*k;i++) {
            if(sum3[i].first > ans) {
                ans = sum3[i].first;
                p[0] = i;
                p[1] = sum3[i].second;
                p[2] = sum2[sum3[i].second].second;
            }
        }
        return p;
    }
};
36915