<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>& v, int k) {
        int n=v.size();
       vector<int>x;
       vector<int>ans;
       for(int i=0;i<n-k+1;i++){
        int sum=0;
        for(int j=i;j<i+k;j++){
            sum+=v[j];
        }
        x.push_back(sum);
       }
       int m=x.size();
       vector<vector<int>>dp(m+1,vector<int>(4,-1));
       for(int i=0;i<=m;i++){
        dp[i][0]=0;
       }
       for(int j=0;j<4;j++){
        dp[0][j]=0;
       }
       for(int j=1;j<4;j++){
        for(int i=k*(j-1)+1;i<=m;i++){
            if(i<k){
                dp[i][j]=max(x[i-1],dp[i-1][j]);
                continue;
            }
            dp[i][j]=max(dp[i-1][j],x[i-1]+dp[i-k][j-1]);
        }
       }
       int mx=dp[m][3];
       int j=3;
       while(j>0){
        for(int i=1;i<=m;i++){
            if(dp[i][j]==mx){
                ans.push_back(i-1);
                mx-=x[i-1];
                break;
            }
        }
        j--;
       }
       reverse(ans.begin(),ans.end());
        return ans;
    }
};
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