<p>You are given an integer array <code>nums</code> and an integer <code>k</code>. Find the maximum subarray sum of all the subarrays of <code>nums</code> that meet the following conditions:</p>

<ul>
	<li>The length of the subarray is <code>k</code>, and</li>
	<li>All the elements of the subarray are <strong>distinct</strong>.</li>
</ul>

<p>Return <em>the maximum subarray sum of all the subarrays that meet the conditions</em><em>.</em> If no subarray meets the conditions, return <code>0</code>.</p>

<p><em>A <strong>subarray</strong> is a contiguous non-empty sequence of elements within an array.</em></p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,5,4,2,9,9,9], k = 3
<strong>Output:</strong> 15
<strong>Explanation:</strong> The subarrays of nums with length 3 are:
- [1,5,4] which meets the requirements and has a sum of 10.
- [5,4,2] which meets the requirements and has a sum of 11.
- [4,2,9] which meets the requirements and has a sum of 15.
- [2,9,9] which does not meet the requirements because the element 9 is repeated.
- [9,9,9] which does not meet the requirements because the element 9 is repeated.
We return 15 because it is the maximum subarray sum of all the subarrays that meet the conditions
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [4,4,4], k = 3
<strong>Output:</strong> 0
<strong>Explanation:</strong> The subarrays of nums with length 3 are:
- [4,4,4] which does not meet the requirements because the element 4 is repeated.
We return 0 because no subarrays meet the conditions.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= k &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution:
    def maximumSubarraySum(self, nums: List[int], k: int) -> int:
        '''
        sliding window

             l
                 r
        [1,1,1,7,8,9]
        0 1 2 3 4 5 6

        {1, 7, 8}
          l
              r
        5 3 3 1 1 

        set = 5, 3, 2
        len = 5 

        7
        
        '''
        left = 0
        N = len(nums)
        maxSum = 0
        curSum = 0
        numSet = set(nums)

        # if len(numSet) + k > len(nums):
        #     return 0

        numSet = set()

        for right in range(N):
            while nums[right] in numSet:
                curSum -= nums[left]
                numSet.remove(nums[left])
                left += 1
                
            curSum += nums[right]
            numSet.add(nums[right])

            # valid window
            if right - left + 1 == k:
                maxSum = max(maxSum, curSum)
                curSum -= nums[left]
                numSet.remove(nums[left])
                left += 1

        return maxSum


        
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