README_EN.md

740. Delete and Earn

中文文档

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.

 

Constraints:

• 1 <= nums.length <= 2 * 104
• 1 <= nums[i] <= 104

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

class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        mx = float('-inf')
        for num in nums:
            mx = max(mx, num)
        total = [0] * (mx + 1)
        for num in nums:
            total[num] += num
        first = total[0]
        second = max(total[0], total[1])
        for i in range(2, mx + 1):
            cur = max(first + total[i], second)
            first = second
            second = cur
        return second

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]);
    }
}

Go

func deleteAndEarn(nums []int) int {

	max := func(x, y int) int {
		if x > y {
			return x
		}
		return y
	}

	mx := math.MinInt32
	for _, num := range nums {
		mx = max(mx, num)
	}
	total := make([]int, mx+1)
	for _, num := range nums {
		total[num] += num
	}
	first := total[0]
	second := max(total[0], total[1])
	for i := 2; i <= mx; i++ {
		cur := max(first+total[i], second)
		first = second
		second = cur
	}
	return second
}

C++

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> vals(10010);
        for (int& num : nums) {
            vals[num] += num;
        }
        return rob(vals);
    }

    int rob(vector<int>& nums) {
        int a = 0, b = nums[0];
        for (int i = 1; i < nums.size(); ++i) {
            int c = max(nums[i] + a, b);
            a = b;
            b = c;
        }
        return b;
    }
};

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