<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>

#include <vector>
#include <numeric>

using namespace std;

class Solution {
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> sum(n + 1, 0);
        
        // Префиксные суммы
        for (int i = 0; i < n; ++i) {
            sum[i + 1] = sum[i] + nums[i];
        }

        vector<int> left(n, 0), right(n, n - k);

        // Вычисляем лучший левый интервал
        for (int i = k, total = sum[k] - sum[0]; i < n; ++i) {
            if (sum[i + 1] - sum[i + 1 - k] > total) {
                left[i] = i + 1 - k;
                total = sum[i + 1] - sum[i + 1 - k];
            } else {
                left[i] = left[i - 1];
            }
        }

        // Вычисляем лучший правый интервал
        for (int i = n - k - 1, total = sum[n] - sum[n - k]; i >= 0; --i) {
            if (sum[i + k] - sum[i] >= total) {
                right[i] = i;
                total = sum[i + k] - sum[i];
            } else {
                right[i] = right[i + 1];
            }
        }

        // Поиск трех непересекающихся интервалов с максимальной суммой
        vector<int> result(3, -1);
        for (int i = k; i <= n - 2 * k; ++i) {
            int l = left[i - 1], r = right[i + k];
            if (result[0] == -1 || 
                sum[l + k] - sum[l] + sum[i + k] - sum[i] + sum[r + k] - sum[r] > 
                sum[result[0] + k] - sum[result[0]] + sum[result[1] + k] - sum[result[1]] + sum[result[2] + k] - sum[result[2]]) {
                result = {l, i, r};
            }
        }

        return result;
    }
};

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