# 1695. Maximum Erasure ValueSolution

[**1695. Maximum Erasure Value**](https://leetcode.com/problems/maximum-erasure-value/)

You are given an array of positive integers `nums` and want to erase a subarray containing **unique elements**. The **score** you get by erasing the subarray is equal to the **sum** of its elements.

Return *the **maximum score** you can get by erasing **exactly one** subarray.*

An array `b` is called to be a subarray of `a` if it forms a contiguous subsequence of `a`, that is, if it is equal to `a[l],a[l+1],...,a[r]` for some `(l,r)`.

**Example 1:**

```
Input: nums = [4,2,4,5,6]
Output: 17
Explanation: The optimal subarray here is [2,4,5,6].
```

**Example 2:**

```
Input: nums = [5,2,1,2,5,2,1,2,5]
Output: 8
Explanation: The optimal subarray here is [5,2,1] or [1,2,5].
```

**Constraints:**

* `1 <= nums.length <= 105`
* `1 <= nums[i] <= 104`

**My Solutions:**

这道题需要找在nums中的一个subarray which only contains unique elements. 用sliding window记录满足条件的array，l代表sliding window的左端，r代表右端。

* 如果当前数字n没有见过，就把此数字和他的index记录在seen里，更新sum和ans。
* 如果此数字见过，比如example2中走到\[5,2,1,2]，此时index=3, 数字2已经在index=1时出现过。把从指针l所在的数字（index=0）到这个数字之前所在位置（index=1）中间位置的所有数字从seen里删掉，此时seen里只剩下数字1。更新sum（从8变成1）和l（变成数字1所在的index=2）。

```
class Solution {
    public int maximumUniqueSubarray(int[] nums) {
        // l代表sliding window的最左端
        int l = 0, sum = 0, ans = 0;
        
        // 用map记录见过的数字和其index
        Map<Integer, Integer> seen = new HashMap<>();
    
        for (int r = 0; r < nums.length; r++){
            int n = nums[r];
            if (seen.containsKey(n)) {
                int index = seen.get(n);
                while (l <= index) {
                    seen.remove(nums[l]);
                    sum -= nums[l];
                    l++;
                }
            }
            seen.put(n, r);
            sum += n;
            ans = Math.max(ans,sum);
        }
        return ans;
    }
}
```