Maximum Erasure Value

Description

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

Code

和 Longest Substring Without Repeating Characters 很類似。

要注意初始 int i = 0 和 if(mp[nums[j]] >= i) 的設定。在 Longest Substring Without Repeating Characters 中的設定是 int i = -1 和 if(mp[nums[j]] > i) （設定成 if(mp[nums[j]] >= i) 也會 AC）。

不論如何，重點都是要能順利正確的更新 i 的值。

Time Complexity: $O(n)$, Space Complexity: $O(n)$

class Solution {
public:
    int maximumUniqueSubarray(vector<int>& nums) {
        int score = 0;
        int maxScore = 0;
        unordered_map<int, int> mp;
        for(int i = 0, j = 0; j < nums.size(); j++) {
            score += nums[j];
 
            int old_i = i;
            if(mp.find(nums[j]) != mp.end()) {
                if(mp[nums[j]] >= i) {
                    i = mp[nums[j]] + 1;
                }
            }
 
            for(int k = old_i; k < i; k++) {
                score -= nums[k];
            }
 
            maxScore = max(maxScore, score);
            mp[nums[j]] = j;
        }
 
        return maxScore;
    }
};

Source

• Maximum Erasure Value - LeetCode