<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:
    vector<int> maxSumOfThreeSubarrays(vector<int>& nums, int k) {
        vector<int> answer(3); // This will store the starting indices of the three subarrays
      
        // Initialize sums of each subarray and maximum sum variables
        int currentSum = 0, firstSubarraySum = 0, secondSubarraySum = 0, thirdSubarraySum = 0;
        int maxFirstSubarraySum = 0, maxFirstSecondSum = 0;
      
        // Indices for tracking the maximum sums for subarrays
        int maxFirstIndex = 0, maxFirstSecondIndex1 = 0, maxFirstSecondIndex2 = 0;
      
        // Loop through the array, starting from the end of the possible third subarray
        for (int i = k * 2; i < nums.size(); ++i) {
            // Increment the sum of each subarray with the next element in the sequence
            firstSubarraySum += nums[i - k * 2];
            secondSubarraySum += nums[i - k];
            thirdSubarraySum += nums[i];
          
            // If we have completed the subarrays (reached full length of k for each)
            if (i >= k * 3 - 1) {
                // Update maximum sum and index for the first subarray if needed
                if (firstSubarraySum > maxFirstSubarraySum) {
                    maxFirstSubarraySum = firstSubarraySum;
                    maxFirstIndex = i - k * 3 + 1;
                }
              
                // Update the sum and indices for the first and second subarray
                if (maxFirstSubarraySum + secondSubarraySum > maxFirstSecondSum) {
                    maxFirstSecondSum = maxFirstSubarraySum + secondSubarraySum;
                    maxFirstSecondIndex1 = maxFirstIndex;
                    maxFirstSecondIndex2 = i - k * 2 + 1;
                }
              
                // Update the total sum and starting indices for all three subarrays
                if (maxFirstSecondSum + thirdSubarraySum > currentSum) {
                    currentSum = maxFirstSecondSum + thirdSubarraySum;
                    answer = {maxFirstSecondIndex1, maxFirstSecondIndex2, i - k + 1};
                }
              
                // Move the window forward by subtracting the outgoing element from each subarray sum
                firstSubarraySum -= nums[i - k * 3 + 1];
                secondSubarraySum -= nums[i - k * 2 + 1];
                thirdSubarraySum -= nums[i - k + 1];
            }
        }
        return answer; // Return the starting indices of the subarrays
    }
};
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