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].

Approach #1: DP. [C++]

class Solution {

public:

    int deleteAndEarn(vector<int>& nums) {

        int len = nums.size();

        if (len == 0) return 0;

        int r = *max_element(nums.begin(), nums.end());

        vector<int> points(r+1, 0);

        for (int num : nums)

            points[num] += num;

        return solve(points);

    }

private:

    int solve(const vector<int>& points) {

        int dp1 = 0, dp2 = 0;

        for (int point : points) {

            int dp = max(dp2 + point, dp1);

            dp2 = dp1;

            dp1 = dp;

        }

        return dp1;

    }

};

Analysis:

If we take nums[i], we can safely take all of its copies. We can't take any of copies of nums[i-1] and nums[i+1], This problem is reduced to 198 House Robber.

Houses[i] has all the copies of num whose value is i.

[3, 4, 2] -> [0, 2, 3, 4], rob([0, 2, 3, 4]) = 6

[2, 2, 3, 3, 3, 4] -> [0, 2*2, 3*3, 4], rob([0, 2*2, 3*3, 4]) = 9

Time complexity: O(n+r) reduction + O(r) solving rob = O(n + r)

Space complexity: O(r).

Reverence:

http://zxi.mytechroad.com/blog/dynamic-programming/leetcode-740-delete-and-earn/