<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:
        result = 0
        values = []
        start=0
        curr_sum=0
        for i in range(len(nums)):
            while nums[i] in values:
                values.pop(0)
                curr_sum-=nums[start]
                start+=1
            values.append(nums[i])
            curr_sum +=nums[i]
            result = max(result,curr_sum)
        return result

        

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