Leetcode 740. 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].

Analysis:
There are several tricks in the problem.
 -- Since each element nums[i] is within the range of [1, 10000], we can use the bucket sort to sort the nums.
 -- For deleting each nums[i], we can delete all same numbers together. So when we do the bucket sort, we can sum up all the repeated numbers in the same bucket. 
 -- Then the problem is a classic backpack problem. 

Define take[10001] and skip[10001], where take[i] means for number i, we delete the number i, the max number of points. skip[i] means we don't delete it.

Then the transit function should be:
take[i] = skip[i - 1] +  values[i]
skip[i] = Math.max(take[i - 1], skip[i - 1]);


Code (Java):

class Solution {
    public int deleteAndEarn(int[] nums) {
        int[] dp = new int[100001];
        
        // step 1: bucket sort
        //
        for (int num : nums) {
            dp[num] += num;
        }
        
        int prevTake = 0;
        int prevSkip = 0;
        
        for (int i = 1; i <= 10000; i++) {
            int currTake = prevSkip + dp[i];
            int currSkip = Math.max(prevSkip, prevTake);
            
            prevTake = currTake;
            prevSkip = currSkip;
        }
        
        return Math.max(prevTake, prevSkip);
    }
}