<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>

const static auto fastio = [] {
  ios::sync_with_stdio(0);
  cin.tie(0);
  return 0;
}();

/*
   dp[i][j] -> maximum sum of j non-overlapping subarrays of size k using the last n-i elements
   dp[i][j] = max((i+1<n ? dp[i + 1][j] : 0), (i+k<n ? dp[i + k][j - 1] : 0) + ks[i]);
 */

class Solution {
  public:
    vector<int> maxSumOfThreeSubarrays(const vector<int> &a, int k) {
      const int n = (int)a.size();
      vector<int> ks(n-k+1);
      ks[n-k] = accumulate(a.begin() + n - k, a.end(), 0);
      for (int i = n-k-1; i >= 0; i--) {
        ks[i] = a[i] + ks[i + 1] - a[i+k];
      }
      // cout << "ks: ";
      // for (int v : ks) cout << v << ' ';
      // cout << '\n';
      vector<vector<int>> dp(n, vector<int>(4));
      for (int j = 1; j <= 3; j++) {
        for (int i = n-j*k; i >= 0; i--) {
          dp[i][j] = max((i+1<n ? dp[i + 1][j] : 0), (i+k<n ? dp[i + k][j - 1] : 0) + ks[i]);
        }
      }
      // for (int i = 0; i < n; i++) {
      //   for (int j = 0; j <= 3; j++) {
      //     cout << dp[i][j] << " \n"[j == 3];
      //   }
      // }
      int st{};
      vector<int> res(3);
      for (int i = 0; i < 3; i++) {
        int j = st;
        while (j+1 < n && dp[j][3-i] == dp[j+1][3-i]) j++;
        res[i] = st; while (ks[res[i]]^ks[j]) res[i]++;
        st = res[i] + k;
      }
      return res;
    }
};
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