<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 l=0, r=0; //define left and right pointers for sliding windows approach
        
        vector<int> sum_arr; //sum_arr stores sum of element of nums at index i;
        sum_arr.push_back(0); //initially l and r at 0 so to consider element at index 0 we need to take 0 at index 0
        
        int sum=0, score=nums[0];
        for(auto it:nums) //storing sum into sum_arr
        {
            sum += it;
            sum_arr.push_back(sum);
        }
        
        unordered_map<int, int> mp; //stores element and their index
        
        while(r<nums.size()) //iterate till r pointer is lesser than nums' size 
        {
            if(mp.find(nums[r]) != mp.end()) l=max(l, mp[nums[r]]+1); //when repeated element come we move left pointer of the right of repeated element
            mp[nums[r]] = r; //assigning index to the element
            score = max(score, sum_arr[r+1]-sum_arr[l]); //get max score
            r++; //move r by 1 
        }
        return score;
        
    }
};
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