<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 solve(vector<int>& nums, int &k, int pos, int count, int &n, vector<vector<int>> &dp, vector<vector<char>> &dp1, vector<int> &prefix){
        if(count == 0 || pos > n-k)      return 0;

        if(dp[pos][count] != -1)        return dp[pos][count];

        int sum = prefix[pos+k-1] - prefix[pos] + nums[pos+k-1];

        int t = sum + solve(nums, k, pos+k, count-1, n, dp, dp1, prefix);
        int nt = solve(nums, k, pos+1, count, n, dp, dp1, prefix);
        if(t>=nt){
            dp1[pos][count] = 'T';
        }
        else{
            dp1[pos][count] = 'N';
        }
        return dp[pos][count] = max(t, nt);

    }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        vector<int> prefix(n, 0);
        for(int i=1;i<n;i++)        prefix[i] = prefix[i-1] + nums[i-1];

        vector<vector<int>> dp(n+1, vector<int>(4, -1));
        vector<vector<char>> dp1(n+1, vector<char>(4, '#'));
        int sum = solve(nums, k, 0, 3, n, dp, dp1, prefix);

        vector<int> ans;
        int prev = 0;
        for(int i=3;i>0;i--){
            for(int pos=prev;pos<n;pos++){
                if(dp[pos][i] == sum && dp1[pos][i] == 'T'){
                    ans.push_back(pos);
                    prev = pos+k;
                    
                    sum -= prefix[pos+k-1] - prefix[pos] + nums[pos+k-1];
                    break;
                }
            }
        }

        return ans;
    }
};
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