<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> maxes;
        int sum = 0;
        for(int i=0; i<k; i++) {
            sum += nums[i];
        }
        maxes.push_back(sum);
        for(int i=1; i<=n-k; i++) {
            sum -= nums[i-1];
            sum += nums[i+k-1];
            maxes.push_back(sum);
        }

        vector<vector<int>> dp(3, vector<int>(n, 0));
        vector<vector<int>> prev(3, vector<int>(n, -1));

        dp[0][0] = maxes[0];
        prev[0][0] = 0;
        int max = dp[0][0];
        for (int i = 1; i + k - 1 < n - 2 * k; i++) {
            if (maxes[i] > max) {
                max = maxes[i];
                dp[0][i] = maxes[i];
                prev[0][i] = i;
            } else {
                dp[0][i] = dp[0][i - 1];
                prev[0][i] = prev[0][i - 1];
            }
        }

        max = 0;
        for (int i = k; i + k - 1 < n - k; i++) {
            if (maxes[i] + dp[0][i - k] > max) {
                max = maxes[i] + dp[0][i - k];
                dp[1][i] = maxes[i] + dp[0][i - k];
                prev[1][i] = i - k;
            } else {
                dp[1][i] = dp[1][i - 1];
                prev[1][i] = prev[1][i - 1];
            }
        }

        max = 0;
        for (int i = 2 * k; i + k - 1 < n; i++) {
            if (maxes[i] + dp[1][i - k] > max) {
                max = maxes[i] + dp[1][i - k];
                dp[2][i] = maxes[i] + dp[1][i - k];
                prev[2][i] = i - k;
            } else {
                dp[2][i] = dp[2][i - 1];
                prev[2][i] = prev[2][i - 1];
            }
        }

        vector<int> answer(3, 0);
        max = 0;
        int maxIndex = -1;
        for (int i = 2 * k; i + k - 1 < n; i++) {
            if (dp[2][i] > max) {
                maxIndex = i;
                max = dp[2][i];
            }
        }

        answer[2] = maxIndex;

        int i = prev[2][answer[2]];
        while(i-1 >= 0 && prev[1][i] == prev[1][i-1]) {
            i--;
        }
        answer[1] = i;

        i = prev[1][answer[1]];
        while(i-1 >= 0 && prev[0][i] == prev[0][i-1]) {
            i--;
        }
        answer[0] = i;



        return answer;
    }
};
28644