<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:
    def maxSumOfThreeSubarrays(self, nums: List[int], k: int) -> List[int]:
        # W = [] #array of sums of windows
        # curr_sum = 0
        # for i, x in enumerate(nums):
        #     curr_sum += x
        #     if i >= k: 
        #         curr_sum -= nums[i - k]
        #     if i >= k - 1: 
        #         W.append(curr_sum)

        # left = [0] * len(W)
        # best = 0
        # for i in range(len(W)):
        #     if W[i] > W[best]:
        #         best = i
        #     left[i] = best

        # right = [0] * len(W)
        # best = len(W) - 1
        # for i in range(len(W) - 1, -1, -1):
        #     if W[i] >= W[best]:
        #         best = i
        #     right[i] = best

        # ans = None
        # for j in range(k, len(W) - k):
        #     i, l = left[j - k], right[j + k]
        #     if ans is None or (W[i] + W[j] + W[l] > W[ans[0]] + W[ans[1]] + W[ans[2]]):
        #         ans = i, j, l
        # return ans

        # Step 1: Calculate the sum of all subarrays of length k using sliding window
        n = len(nums)  # Length of the input array
        sum_k = [0] * (n - k + 1)  # Array to store sums of subarrays of length k
        curr_sum = 0  # Variable to keep track of the current window's sum

        # Loop through the nums array and calculate subarray sums of length k
        for i in range(n):
            curr_sum += nums[i]  # Add current number to the running sum
            if i >= k:  # If window size exceeds k, subtract the element that's leaving the window
                curr_sum -= nums[i - k]
            if i >= k - 1:  # When we have a full window (size k), store the sum in sum_k
                sum_k[i - k + 1] = curr_sum
        
        # Step 2: Calculate the best left subarray sum for each index
        left = [0] * len(sum_k)  # left[i] will store the index of the max sum subarray on the left up to index i
        best_left = 0  # Track the best left subarray index
        for i in range(len(sum_k)):
            if sum_k[i] > sum_k[best_left]:  # If we find a better subarray sum, update best_left
                best_left = i
            left[i] = best_left  # Store the best left subarray index up to position i

        # Step 3: Calculate the best right subarray sum for each index
        right = [0] * len(sum_k)  # right[i] will store the index of the max sum subarray on the right from index i
        best_right = len(sum_k) - 1  # Track the best right subarray index
        for i in range(len(sum_k) - 1, -1, -1):  # Iterate from right to left
            if sum_k[i] >= sum_k[best_right]:  # If we find a better or equal subarray sum, update best_right
                best_right = i
            right[i] = best_right  # Store the best right subarray index from position i to the end
        
        # Step 4: Find the maximum sum by testing all possible middle subarrays
        max_sum = 0  # Variable to store the maximum sum found
        result = []  # List to store the indices of the three subarrays with the maximum sum
        # The middle subarray starts from index j, iterate over possible middle subarrays
        for j in range(k, len(sum_k) - k):  
            l = left[j - k]  # Best left subarray before the middle subarray
            r = right[j + k]  # Best right subarray after the middle subarray
            current_sum = sum_k[l] + sum_k[j] + sum_k[r]  # Total sum of the three subarrays
            if current_sum > max_sum:  # If the current sum is greater than max_sum, update max_sum and result
                max_sum = current_sum
                result = [l, j, r]  # Store the indices of the three subarrays
        
        return result  # Return the indices of the three subarrays with the maximum sum


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