<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 n;vector<int> ans;
    int rec(int level,int take,vector<int>& nums,int k,vector<vector<int>>& dp,vector<int>& pre){
        if(level==n){
            if(take==3) return 0;
            return INT_MIN;
        }
        if(dp[level][take]!=-1) return dp[level][take];
        int ans=rec(level+1,take,nums,k,dp,pre);
        if(level+k-1<n && take<3) ans=max(ans,rec(level+k,take+1,nums,k,dp,pre)+pre[level+k-1]+nums[level+k-1]-pre[level]);
        return dp[level][take]=ans;
    }
    void print(int level,int take,vector<int>&nums,int k,vector<vector<int>>& dp,vector<int>& pre){
        if(level==n) return;
        int takke=INT_MIN,nottake=rec(level+1,take,nums,k,dp,pre);
        if(take<3 & level+k<n) takke=rec(level+k,take+1,nums,k,dp,pre)+pre[level+k-1]+nums[level+k-1]-pre[level];
        if(takke>=nottake){
           ans.push_back(level);
            print(level+k,take+1,nums,k,dp,pre);
        }
        else print(level+1,take,nums,k,dp,pre);
        // cout<<level<<" "<<takke<<" "<<nottake<<endl;
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        n=nums.size();
        vector<int> pre(n); vector<vector<int>> dp(n,vector<int>(5,-1));
        pre[0]=0;
        for(int i=1;i<n;i++) pre[i]=nums[i-1]+pre[i-1];
        cout<<rec(0,0,nums,k,dp,pre)<<endl;
        print(0,0,nums,k,dp,pre);
        
        return ans;
    }
};
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