<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 f(int ind, int cnt, vector<vector<int>>&dp, vector<int>&nums, int k){
        if(cnt == 0) return 0;
        if(nums.size()-ind < cnt*k) return INT_MIN;
        if(dp[ind][cnt]==0){
            int notake = f(ind+1, cnt, dp, nums, k);
            int take = f(ind+k, cnt-1, dp, nums, k);
            if(take != INT_MIN)
                take += accumulate(nums.begin()+ind, nums.begin()+ind+k, 0);     
            dp[ind][cnt] = max(take, notake);
        }
        return dp[ind][cnt];
    }

    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n=nums.size();
        vector<vector<int>>dp(n+10, vector<int>(4,0));
        f(0,3,dp,nums,k);
        int i=0, j=3;
        vector<int>ans;
        while(j!=0){
            if(i+k == nums.size() || dp[i][j] == dp[i+k][j-1] + accumulate(nums.begin()+i, nums.begin()+i+k,0)){
                ans.push_back(i);
                i=i+k;
                j--;
            }
            else
                i++;
        }
        return ans;
    }
};


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