<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:
    int sum=0;
    int solve(vector<int>&pre,int ind, int k, int ct,vector<vector<int>>&memo){
        int n=pre.size()-1;
        if(ct==0 || ind > n-k) {
            return 0;
        }

        if(memo[ind][ct]) return memo[ind][ct];

        int take = -1e9;
        if(ind + k <= n){
            take = pre[ind+k] - pre[ind] + solve(pre,ind+k,k,ct-1,memo);
        }

        int nottake = solve(pre,ind+1,k,ct,memo);

        

        return memo[ind][ct] = max(take,nottake);
    }

    vector<int> f(vector<int>&pre, int ind,int k, int ct, vector<int>index, vector<vector<int>>&memo,int maxi){
        int n=pre.size()-1;
        if(ct==0 && !maxi){
            return index;
        }

        if(ct==0 || ind>n-k || memo[ind][ct]!=maxi){
            int m = 1e9;
            return {m,m,m};
        }

        int sum = pre[ind+k]-pre[ind];
        index.push_back(ind);
        auto take = f(pre,ind+k,k,ct-1,index,memo,maxi-sum);
        index.pop_back();
        auto nottake = f(pre,ind+1,k,ct,index,memo,maxi);

        return min(take,nottake);
    }

    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n=nums.size();
        vector<int> pre(n+1,0);
        vector memo(n + 1, vector<int>(4));


        for(int i=0;i<n;i++){
            pre[i+1] = nums[i] + pre[i];
        }

        vector<int>ans(3,0),temp;
        int maxi=solve(pre,0,k,3,memo);

        return f(pre,0,k,3,{},memo,maxi);
    }
};
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