740. Delete and Earn

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.

 

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 * 104
• 1 <= nums[i] <= 104

Solution

delete-and-earn.py

class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        buckets = [0] * 10001

        for x in nums:
            buckets[x] += x

        dp = [0] * 10001
        dp[0] = buckets[0]
        dp[1] = buckets[1]

        for i in range(2, 10001):
            dp[i] = max(dp[i - 1], dp[i - 2] + buckets[i])

        return dp[-1]

delete-and-earn.cpp

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> buckets(10001,0);

        for (int i = 0; i < nums.size(); i++){
            buckets[nums[i]] += nums[i];
        }

        vector<int> dp(10001,0);
        dp[0] = buckets[0];
        dp[1] = buckets[1];

        for (int i = 2; i < buckets.size(); i++){
            dp[i] = max(dp[i-1], buckets[i]+dp[i-2]);
        }

        return dp[10000];
    }
};