<p>You are given an array of positive integers <code>nums</code> and want to erase a subarray containing&nbsp;<strong>unique elements</strong>. The <strong>score</strong> you get by erasing the subarray is equal to the <strong>sum</strong> of its elements.</p>

<p>Return <em>the <strong>maximum score</strong> you can get by erasing <strong>exactly one</strong> subarray.</em></p>

<p>An array <code>b</code> is called to be a <span class="tex-font-style-it">subarray</span> of <code>a</code> if it forms a contiguous subsequence of <code>a</code>, that is, if it is equal to <code>a[l],a[l+1],...,a[r]</code> for some <code>(l,r)</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [4,2,4,5,6]
<strong>Output:</strong> 17
<strong>Explanation:</strong> The optimal subarray here is [2,4,5,6].
</pre>

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

<pre>
<strong>Input:</strong> nums = [5,2,1,2,5,2,1,2,5]
<strong>Output:</strong> 8
<strong>Explanation:</strong> The optimal subarray here is [5,2,1] or [1,2,5].
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def maximumUniqueSubarray(self, nums: List[int]) -> int:
        count = {}
        l, res = 0, 0
        currSum = 0
        for r in range(len(nums)): 
            count[nums[r]] = count.get(nums[r], 0) + 1
            currSum += nums[r]
            while nums[r] in count and count[nums[r]] > 1:
                count[nums[l]] -= 1 #guaranteed that count[nums[l]] exists since we passed it, its a window    
                currSum -= nums[l]
                if count[nums[l]] == 0:
                    count.pop(nums[l])
                l += 1
            res = max(res, currSum)
        return res
1321