740. Delete and Earn

740. Delete and Earn (Medium)

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].

 

Related Topics:
Dynamic Programming

Similar Questions:

• House Robber (Easy)

Solution 1. DP

Firstly, to avoid duplicate, store the data in a map from the number to its count.

Let dp[i] be the max point you can get at point i.

If num != prevNum + 1, we can freely pick num, then dp[i] = dp[i-1] + num * count.

Otherwise, if we don't pick num, dp[i] = dp[i-1].

Otherwise, we pick num, dp[i] = dp[i-2] + num * count.

So in sum:

dp[i] = num == prevNum ? max(dp[i-1], dp[i-2] + num * count) : (dp[i-1] + num * count)

// OJ: https://leetcode.com/problems/delete-and-earn/
// Author: github.com/lzl124631x
// Time: O(NlogN)
// Space: O(N)
class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        map<int, int> m;
        for (int n : nums) m[n]++;
        int prev = 0, prev2 = 0, num = INT_MIN;
        for (auto &p : m) {
            int cur = p.first == num + 1 ? max(prev, prev2 + p.first * p.second) : (prev + p.first * p.second);
            prev2 = prev;
            prev = cur;
            num = p.first;
        }
        return prev;
    }
};