<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, vector<int>& prefixSum, int k, vector<vector<int>>& dp, vector<vector<int>>& choice, int i, int rem) {
        if (rem == 0)
         return 0;  
         if (i + rem * k > nums.size())
          return 0; 
         if (dp[i][rem] != -1)
            return dp[i][rem];
          int nt= 0;
          nt=solve(nums, prefixSum, k, dp, choice, i + 1, rem);
          choice[i][rem] = 0;
          int mx=nt;
           if (i + k <= nums.size()) {
            int s = prefixSum[i + k] - prefixSum[i];
            int t=s+solve(nums,prefixSum,k,dp,choice,i+k,rem-1);
             if (t > nt || (t == nt && choice[i][rem] == 0)) 
            {
               mx=t;
               choice[i][rem]=1;
            }
           }
          return dp[i][rem]=mx;
           }
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        int n = nums.size();
        if (n < 3 * k) return {};
        vector<int> prefixSum(n + 1, 0);
        vector<vector<int>> dp(n + 1, vector<int>(4, -1));
         for (int i = 0; i < n; ++i) {
            prefixSum[i + 1] = prefixSum[i] + nums[i];
        }
         vector<vector<int>> choice(n + 1, vector<int>(4, -1)); 
         solve(nums, prefixSum, k, dp, choice, 0, 3);
          vector<int> indices;
        int index = 0;
        int rem = 3;
        
        while (rem> 0 && index < n) {
            if (choice[index][rem] == 1) {
                indices.push_back(index);
                index += k;
                --rem;
            } else {
                ++index;
            }
        } 
     
   return indices;
    }
};
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