<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 help(int i,vector<int>& window,int k,int cnt,vector<vector<int>>& dp){
        int n=window.size();
        if(i>=n){
            return 0;
        }
        if(cnt==3){
            return 0;
        }
        if(dp[i][cnt]!=-1){
            return dp[i][cnt];
        }
        int take=window[i]+help(i+k,window,k,cnt+1,dp);
        int notTake=help(i+1,window,k,cnt,dp);
        return dp[i][cnt]=max(take,notTake);
    }
    void dfs(int i,int cnt,vector<int>& path,int k,vector<int>& window,vector<vector<int>>& dp){
        if(cnt==3){
            return;
        }
        int take=window[i]+help(i+k,window,k,cnt+1,dp);
        int notTake=help(i+1,window,k,cnt,dp);
        if(take>=notTake){
            path.push_back(i);
            dfs(i+k,cnt+1,path,k,window,dp);
        }
        else{
            dfs(i+1,cnt,path,k,window,dp);
        }
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& v, int k) {
        vector<int> pre;
        pre.push_back(0);
        for(int i=0;i<v.size();i++){
            pre.push_back(pre.back()+v[i]);
        }
        int n=pre.size();
        vector<int> window;
        for(int i=0;i+k<n;i++){
            window.push_back(pre[i+k]-pre[i]);
        }
        n=window.size();
        vector<vector<int>> dp(n,vector<int>(4,-1));
        cout<<help(0,window,k,0,dp)<<endl;
        vector<int> path;
        dfs(0,0,path,k,window,dp);
        return path;
    }
};
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