<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 {
public:
    int maximumUniqueSubarray(vector<int>& nums) {
        //Candidate - keeps track of highest sum
        int candidateSum = nums[0];
        //Prefix array
        vector<int> prefix(nums.size());
        prefix[0] = nums[0];
        //Unique values set
        unordered_set<int> uniqueVals;
        uniqueVals.insert(nums[0]);

        //Generate prefix sum
        for (int i=1; i<nums.size(); i++) {
            prefix[i] = nums[i] + prefix[i-1];
        }

        //Sliding window - keep track of unique subarrays
        int i=0;
        int j=0;
        int localSum;
        while (j+1<nums.size()) {
            //Attempt to increase the window
            j++;
            //If we run into a previously discovered value
            if (uniqueVals.find(nums[j]) != uniqueVals.end()) {
                //Move the window up until we pass the duplicate value just landed
                while (nums[i] != nums[j] && i<j) {
                    uniqueVals.erase(nums[i]);
                    i++;
                }
                if (i!=j) {
                    i++;
                }
            }
            //If its a new unique value, add to unique vals
            else {
                uniqueVals.insert(nums[j]);
            }

            //Calculate sum of current subarray
            if (i>0) {
                localSum = prefix[j] - prefix[i-1];
            }
            else {
                localSum = prefix[j];
            }

            //If it's the highest sum, add it as the candiate
            candidateSum = max(localSum, candidateSum);
        }

        return candidateSum;
    }
};
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