# 740. Delete and Earn

[Link](https://leetcode.com/problems/delete-and-earn/)

Given an array `nums` of integers, you can perform operations on the array.

In each operation, you pick any `nums[i]` and delete it to earn `nums[i]` points. After, you must delete **every** element equal to `nums[i] - 1` or `nums[i] + 1`.

You start with 0 points. Return the maximum number of points you can earn by applying such operations.

**Example 1:**

```
Input: nums = [3, 4, 2]
Output: 6
Explanation: 
Delete 4 to earn 4 points, consequently 3 is also deleted.
Then, delete 2 to earn 2 points. 6 total points are earned.
```

**Example 2:**

```
Input: nums = [2, 2, 3, 3, 3, 4]
Output: 9
Explanation: 
Delete 3 to earn 3 points, deleting both 2's and the 4.
Then, delete 3 again to earn 3 points, and 3 again to earn 3 points.
9 total points are earned.
```

**Note:**

* The length of `nums` is at most `20000`.
* Each element `nums[i]` is an integer in the range `[1, 10000]`.

#### Solution

1. If we sort all the numbers into `buckets` indexed by these numbers, this is essentially asking you to repetitively take an bucket while giving up the 2 buckets next to it. (the range of these numbers is \[1, 10000])
2. The optimal final result can be derived by keep updating 2 variables `skip_i`, `take_i`, which stands for:\
   `skip_i` : the best result for sub-problem of first `(i+1)` buckets from `0` to `i`, while you **skip** the `i`th bucket.\
   `take_i` : the best result for sub-problem of first `(i+1)` buckets from `0` to `i`, while you **take** the `i`th bucket.
3. DP formula:\
   `take[i] = skip[i-1] + values[i];`\
   `skip[i] = Math.max(skip[i-1], take[i-1]);`

```java
  public int deleteAndEarn(int[] nums) {
        int[] buckets = new int[10001];
        
        for(int num : nums){
            buckets[num] += num;
        }
        // dp[i] means the max value that I can get at the buckets[i] 
        int[] dp = new int[10001];
        dp[0] = buckets[0];
        dp[1] = buckets[1];
        
        for(int i = 2; i < 10001; i++){
            //not skip this time, or skip
            dp[i] = Math.max(dp[i-2]+buckets[i], dp[i-1]);
            
        }
        return dp[10000];
        
    }
```
