<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) {
        const int n = nums.size();
        vector<vector<int>> dp(n+2, vector<int> (4, -1e8));
        vector<int> preffix(n+2, 0);
        preffix[0] = nums[0];
        for(int i=1;i<n;i++) preffix[i] = preffix[i-1] + nums[i]; 
        
        dp[n][3] = 0;

        for(int pos = n - 1; pos >= 0; pos--){
            for(int k = 0; k <= 3; k++){
                int take = k + 1 <= 3 && pos + K <= n
                        ? dp[pos + K][k + 1] + preffix[pos + K - 1] - (pos - 1 >= 0 ? preffix[pos - 1] : 0) 
                        : -1e8;
                int notTaken = dp[pos+1][k];
                dp[pos][k] = max(take, notTaken);
            }
        }

        vector<int> resp;
        int auxK = 0;
        for(int pos=0;pos<=n-K && auxK < 3;){
            int take = dp[pos+K][auxK + 1] + preffix[pos + K - 1] - (pos - 1 >= 0 ? preffix[pos - 1] : 0);
            int notTaken = dp[pos+1][auxK];
            if(take >= notTaken){
                resp.push_back(pos);
                pos += K;
                auxK++;
            }else{
                pos++;
            }
        }

        return resp;
    }
};
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