<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) {
        int sum = 0;
        int maxSum = 0;
        unordered_set<int> mp;
        vector<int> temp = {nums[0]};
        for(int i = 1; i < nums.size(); i++){
            temp.push_back(temp[i-1]+nums[i]);
        }
        int left = 0;
        int right = 0;
        while(right < nums.size()){
            while(mp.find(nums[right]) != mp.end()){
                mp.erase(nums[left]);
                left++;
            }
            maxSum = max(maxSum, temp[right]-temp[left]+nums[left]);
            mp.insert(nums[right]);
            right++;
        }
        /*for(int num : nums){
            sum += num;
            while(mp.find(num) != mp.end()){
                sum -= nums[left];
                mp.erase(nums[left]);
                left++;
            }
            maxSum = max(maxSum, sum);
            mp.insert(num);
        }*/
        return maxSum;
    }
};
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