<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>

#define ll long long
class Solution {
public: vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
	const int inf=1e9;
    int n=nums.size(),m=3;
    vector<vector<int>>dp(1+n,vector<int>(1+m,-inf));
    vector<int>pSum(1+n);
    for(int i=0;i<=n;i++)
        dp[i][0]=0;
    for(int i=1;i<=n;i++)
    {
        pSum[i]=pSum[i-1]+nums[i-1];
        for(int j=1;j<=m;j++)
        {
            dp[i][j]=dp[i-1][j];
            if(i-k>=0)
                dp[i][j]=max(dp[i][j],dp[i-k][j-1]+pSum[i]-pSum[i-k]);
        }
    }
    vector<int>ans;
    int i=n,j=m;
    while(j>0)
    {
        if(dp[i][j]==dp[i-1][j])
            i--;
        else
        {
            ans.push_back(i-k);
            i-=k;
            j--;
        }
    }
    reverse(ans.begin(),ans.end());
    return ans;
}
	
};
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