<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 {
    vector<int> ans;
private:
    int f(vector<int>& nums, int k, int cnt, int ind,vector<vector<int>>& dp){
        if(cnt==3) return 0;
        if(ind>=nums.size() && cnt!=3) return INT_MIN;
        if(dp[ind][cnt]!=0) return dp[ind][cnt];
        int not_take=f(nums,k,cnt,ind+1,dp);
        int take=0;
        if(ind+k<=nums.size()){
            take=accumulate(nums.begin()+ind,nums.begin()+ind+k,0)+f(nums,k,cnt+1,ind+k,dp);
        }
        return dp[ind][cnt]=max(take,not_take);
    }
public:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        vector<vector<int>> dp(nums.size()+1,vector<int>(4,0));
        int sum=f(nums,k,0,0,dp);
        //for(auto it:dp){
        //    for(auto i:it) cout<<i<<" ";
        //    cout<<endl;
        //}
        vector<int> res;
        int j=0;
        int i=0;
        while(j!=3){
            if(i+k<=nums.size() && dp[i+k][j+1]==dp[i][j]-accumulate(nums.begin()+i,nums.begin()+i+k,0)){
                res.push_back(i);
                j++;
                i=i+k;
                //if(j==2) res.push_back(i+k);
            }
            else i++;
        }
        return res;
    }
};
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