<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 ;
        map<int,int> m ;
        int count = 0 ;
        int i=0 ; 
        int j=0 ; 
        int maxi = 0 ;
        while(j<nums.size()){
            sum = sum + nums[j] ;
            m[nums[j]]++;
            if(m[nums[j]] > 1){
                count++ ;
            }
            while(count > 0){
                sum = sum - nums[i] ;
                if(m[nums[i]] > 1){
                    m[nums[i]]--;
                    count-- ;
                }else{
                     m[nums[i]]--;
                }
                if(m[nums[i]] == 0){
                    m.erase(nums[i]) ;
                }
                i++;
            }
            maxi = max(maxi ,sum) ;
            j++ ;
        }
        return maxi ;
    }
};
136907