<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:
    vector<int> maxSumOfThreeSubarrays(vector<int>& arr, int k) {

        int n=arr.size();


        vector<int> pref(n+1,0);
        int i,j;
        for(i=1;i<=n;i++){pref[i]=pref[i-1]+arr[i-1];}

        vector<vector<int>> dp(n+1,vector<int>(4,0));

        vector<vector<vector<int>>> dpp(n+1,vector<vector<int>>(4));
        for(i=1;i<=n;i++){
            dp[i][0]=0;
            dp[i][1]=dp[i-1][1];
            dp[i][2]=dp[i-1][2];
            dp[i][3]=dp[i-1][3];
            if(i-k>=0){
            dp[i][1]=max(dp[i-1][1],dp[i-k][0]+pref[i]-pref[i-k]);
            dp[i][2]=max(dp[i-1][2],dp[i-k][1]+pref[i]-pref[i-k]);
            dp[i][3]=max(dp[i-1][3],dp[i-k][2]+pref[i]-pref[i-k]);}


            if(dp[i][1]==dp[i-1][1]){dpp[i][1]=dpp[i-1][1];}
            else{auto v=dpp[i-k][0];v.push_back(i-k);dpp[i][1]=v;}

            if(dp[i][2]==dp[i-1][2]){dpp[i][2]=dpp[i-1][2];}
            else{auto v=dpp[i-k][1];v.push_back(i-k);dpp[i][2]=v;}


            if(dp[i][3]==dp[i-1][3]){dpp[i][3]=dpp[i-1][3];}
            else{auto v=dpp[i-k][2];v.push_back(i-k);dpp[i][3]=v;}


        }
      //  for(auto &i:dpp[n][3]){i--;}
        return dpp[n][3];
vector<int> ans;ans.push_back(dp[n][3]);
        return ans;
    }
};
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