740. Delete and Earn

Problem

You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:

• Pick any nums[i] and delete it to earn nums[i] points. Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.

Return **the *maximum number of points* you can earn by applying the above operation some number of times**.

  Example 1:

Input: nums = [3,4,2]
Output: 6
Explanation: You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.

Example 2:

Input: nums = [2,2,3,3,3,4]
Output: 9
Explanation: You can perform the following operations:
- Delete a 3 to earn 3 points. All 2's and 4's are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.

  Constraints:

• 1 <= nums.length <= 2 * 10^4

• 1 <= nums[i] <= 10^4

Solution (Java)

class Solution {
    public int deleteAndEarn(int[] nums) {
        int[] sum = new int[10001];
        int min = 10001;
        int max = 0;
        for (int num : nums) {
            sum[num] += num;
            min = Math.min(num, min);
            max = Math.max(num, max);
        }

        int[] dp = new int[max + 1];
        dp[min] = sum[min];
        for (int i = min + 1; i <= max; i++) {
            dp[i] = Math.max(dp[i - 1], dp[i - 2] + sum[i]);
        }
        return dp[max];
    }
}

Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).