<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<vector<int>>dp;
    vector<int>pre;
    map<pair<int,int>,int>par;
    int solve(int i,int j,int k,vector<int>& nums)
    {
        if(j>2)
        {
            return 0;
        }
        if(i+k>nums.size())
        {
            return -1e9;
        }
        if(dp[i][j]!=-1)
        {
            return dp[i][j];
        }
        int x,y;
        x=solve(i+1,j,k,nums);
        y=pre[i+k]-pre[i]+solve(i+k,j+1,k,nums);
        if(y>=x)
            par[{i,j}]=1;
        else
            par[{i,j}]=0;
        return dp[i][j]=max(x,y);
    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) 
    {
        int n=nums.size(),x,y;
        pre.resize(n+1,0);
        for(int i=1;i<=n;++i)
        {
            pre[i]=nums[i-1]+pre[i-1];
        }
        dp.resize(n,vector<int>(3,-1));
        int ans=solve(0,0,k,nums);
        // cout<<ans;
        // for(int i=0;i<n;++i)
        // {
        //     for(int j=0;j<3;++j)
        //     cout<<dp[i][j]<<" ";
        //     cout<<'\n';
        // }
        x=0;
        y=0;
        vector<int>sol;
        while(par.find({x,y})!=par.end())
        {
            if(par[{x,y}])
            {
                sol.push_back(x);
                x=x+k;
                y=y+1;
            }
            else
            {
                x=x+1;
                y=y;
            }
            if(sol.size()==3)
            {
                break;
            }
            // cout<<x<<' '<<y<<'\n';
        }
        return sol;
    }
};
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