README_EN.md

740. Delete and Earn

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Description

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:

<li>The length of <code>nums</code> is at most <code>20000</code>.</li>

<li>Each element <code>nums[i]</code> is an integer in the range <code>[1, 10000]</code>.</li>

 

Solutions

Intuition: If we take a number, we will take all of the copies of it.

First calculate the sum of each number as sums, and keep updating two dp arrays: select and nonSelect

• sums[i] represents the sum of elements whose value is i;
• select[i] represents the maximum sum of processing from 0 to i if the number i is selected;
• nonSelect[i] represents the maximum sum of processing from 0 to i if the number i is not selected;

Then we have the following conclusions:

• If i is selected, then i-1 must not be selected;
• If you do not choose i, then i-1 can choose or not, so we choose the larger one;

select[i] = nonSelect[i-1] + sums[i];
nonSelect[i] = Math.max(select[i-1], nonSelect[i-1]);

Python3

Java

class Solution {
    public int deleteAndEarn(int[] nums) {
        if (nums.length == 0) {
            return 0;
        }

        int[] sums = new int[10010];
        int[] select = new int[10010];
        int[] nonSelect = new int[10010];

        int maxV = 0;
        for (int x : nums) {
            sums[x] += x;
            maxV = Math.max(maxV, x);
        }

        for (int i = 1; i <= maxV; i++) {
            select[i] = nonSelect[i - 1] + sums[i];
            nonSelect[i] = Math.max(select[i - 1], nonSelect[i - 1]);
        }
        return Math.max(select[maxV], nonSelect[maxV]);
    }
}

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