<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>& nums, int k) {
        int n=nums.size();
        vector<vector<pair<int,int>>>dp(n,vector<pair<int,int>>(4,{0,-1}));

        vector<int>psum(n);
        psum[0]=nums[0];
        for(int i=1;i<n;i++) psum[i]=nums[i]+psum[i-1];

        for(int i=0;i<n;i++)
        {
            for(int j=1;j<=3;j++)
            {
                if(k*j>(i+1)) continue;
                if(i-1>=0) dp[i][j]=dp[i-1][j];
                if(i-k+1>=0)
                {
                    int sum=psum[i]-((i-k>=0)?(psum[i-k]-dp[i-k][j-1].first):0);
                    if(sum>dp[i][j].first) 
                    {
                        dp[i][j].first=sum;
                        dp[i][j].second=i-k+1;
                    }
                }
                // cout<<i<<" "<<j<<" "<<dp[i][j].first<<" "<<dp[i][j].second<<endl;
            }
        }
       
        int fst=dp[n-1][3].second;
        int snd=dp[fst-1][2].second;
        int trd=dp[snd-1][1].second;
        return {trd,snd,fst};
        
    }
};
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