LeetCode 1186. Delete once to get the sub array maximum and daily question

Problem description

Give you an array of integers , Return one of its   Non empty Subarray （ Continuous element ） After performing an optional delete operation , The maximum sum of elements that can be obtained . let me put it another way , You can choose a sub array from the original array , And you can decide whether to delete an element from it （ It can only be deleted once ）,（ After deleting ） There should be at least one element in the subarray , Then the subarray （ be left over ） The sum of the elements of is the largest of all subarrays .

Be careful , After deleting an element , Subarray Can't be empty .

Example 1：

Input ：arr = [1,-2,0,3]
Output ：4
explain ： We can choose [1, -2, 0, 3], And then delete -2, In this way, I get [1, 0, 3], And the biggest .
Example 2：

Input ：arr = [1,-2,-2,3]
Output ：3
explain ： Let's just pick [3], This is the maximum sum .
Example 3：

Input ：arr = [-1,-1,-1,-1]
Output ：-1
explain ： The resulting subarray cannot be empty , So we can't choose [-1] And delete -1 To get 0.
      We should choose directly [-1], Or choose [-1, -1] Then delete one from it -1.
 

Tips ：

1 <= arr.length <= 105
-104 <= arr[i] <= 104

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/maximum-subarray-sum-with-one-deletion
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Java

class Solution {
    public int maximumSum(int[] arr) {
        int len = arr.length;
        int remain = arr[0];
        int delete = 0;
        int ans = remain;

        for(int i = 1;i < len;i++){
            int n = arr[i];
            if(n >= 0){
                remain = Math.max(0,remain) + n;
                delete = delete + n;
            }else {
                // Delete this time , Or delete the previous element 
                delete = Math.max(remain,delete + n);
                remain = Math.max(0,remain) + n;
            }
            ans = Math.max(ans,remain);
            ans = Math.max(ans,delete);
        }
        return ans;
    }
}