740. Delete and Earn

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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 ith bucket. take_i : the best result for sub-problem of first (i+1) buckets from 0 to i, while you take the ith bucket.

3. DP formula: take[i] = skip[i-1] + values[i]; skip[i] = Math.max(skip[i-1], take[i-1]);

  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];
        
    }