<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 dp[20001][4];
    int check(int i, int limit, int& k, vector<int>& nums, vector<int>& prefix){
        if(limit==0 || i+k-1>=nums.size()) return 0;
        if(dp[i][limit]!=-1) return dp[i][limit];
        return dp[i][limit] = max(check(i+1,limit,k,nums,prefix),prefix[i+k-1]-prefix[i]+nums[i]+check(i+k,limit-1,k,nums,prefix));
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        memset(dp,-1,sizeof(dp));
        vector<int> prefix(nums.size());
        prefix[0] = nums[0];
        for(int i=1;i<nums.size();i++){
            prefix[i] = prefix[i-1]+nums[i];
        }
        int maxSum = check(0,3,k,nums,prefix);
        vector<int> ans;
        int i=0, limit=3;
        while(ans.size()<3){
            int noPick = check(i+1,limit,k,nums,prefix);
            int pick = prefix[i+k-1]-prefix[i]+nums[i]+check(i+k,limit-1,k,nums,prefix);
            if(dp[i][limit]==pick){
                ans.push_back(i);
                i+=k;
                limit--;
            }
            else i++;
        }
        return ans;
    }
};
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