740. Delete and Earn

1. Description

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.

2. Example

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.

3. Constraints

• 1 <= nums.length <= $2 * 10^4$
• 1 <= nums[i] <= $10^4$

4. Solutions

Dynamic Programming

n = nums.size()
Time complexity: O(nlogn)
Space complexity: O(n)

class Solution {
public:
    int deleteAndEarn(vector<int> &nums) {
        map<int, int> num_count;
        for (auto num : nums) {
            ++num_count[num];
        }

        array<array<int, 2>, 2> dp;
        // dp[0]: delete the number
        // dp[1]: don't delete the number
        int last_number = -1;
        for (auto iter : num_count) {
            dp[1][0] = (iter.first - last_number == 1 ? dp[0][1] : max(dp[0][0], dp[0][1])) +
                iter.first * iter.second;
            dp[1][1] = max(dp[0][0], dp[0][1]);

            last_number = iter.first;
            swap(dp[0], dp[1]);
        }

        return max(dp[0][0], dp[0][1]);
    }
};