<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 {
    int dp[4][21000];
    int n;
    vector<int> v;
public:
    int rec(int left,int level,int k){
        if(level==n){
            if(left){
                return -1e9;
            }
            return 0;
        }
        if(!left){
            return 0;
        }
        if(n-level<left*k){
            return -1e9;
        }
        if(dp[left][level]!=-1){
            return dp[left][level];
        }
        int sum=v[k-1+level];
        if(level){
            sum-=v[level-1];
        }
        int ans=rec(left,level+1,k);
        ans=max(ans,sum+rec(left-1,level+k,k));
        return dp[left][level]=ans;
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        n=nums.size();
        v.resize(n);
        v[0]=nums[0];
        for(int i=1;i<n;i++){
            v[i]=nums[i]+v[i-1];
        }
        memset(dp,-1,sizeof(dp));
        vector<int> ans;
        int val=rec(3,0,k);
        int start=0;
        int sum=0;
        for(int i=0;i<3;i++){
            for(int j=start;j+k-1<n;j++){
                int prefix=v[j+k-1];
                if(j){
                    prefix-=v[j-1];
                }
                if(sum+prefix+rec(2-ans.size(),j+k,k)==val){
                    ans.push_back(j);
                    start=j+k;
                    sum+=prefix;
                    break;
                }
            }
        }
        return ans;
    }
};
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