# 1695-maximum-erasure-value


Try it on <a href='https://leetcode.com/problems/1695-maximum-erasure-value'>leetcode</a>

## Description
<div class="description">
<div><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>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>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>
</div>
</div>

## Solution(Python)
```Python
class Solution:
    def maximumUniqueSubarray(self, nums: List[int]) -> int:
        return self.optimal(nums)

    # time complexity: O(n^2)
    # Space complexity: O(n)
    def bruteforce(self, nums: List[int]) -> int:
        n = len(nums)
        max_sum = 0
        seen = set()
        for i in range(n):
            seen.clear()
            cur_sum = 0

            for j in range(i, n):
                if nums[j] in seen:
                    break
                cur_sum += nums[j]
                seen.add(nums[j])
            max_sum = max(max_sum, cur_sum)
        return max_sum

    # time complexity: O(n)
    # Space complexity: O(n)
    def optimal(self, nums: List[int]) -> int:
        n = len(nums)
        max_sum = 0
        cur_sum = 0
        seen = set()
        l = 0
        for num in nums:
            while num in seen:
                seen.remove(nums[l])
                cur_sum -= nums[l]
                l += 1
            cur_sum += num
            seen.add(num)
            max_sum = max(max_sum, cur_sum)
        return max_sum

```