Delete and Earn

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:

• The length of nums is at most 20000.

• Each element nums[i] is an integer in the range [1, 10000].

Solutions

My solution

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        sort(nums.begin(), nums.end());
        int n = nums.size();
        if(n == 0) return 0;
        // can be optimized to use O(1) space.
        vector<int> pick(n);
        vector<int> skip(n);
        pick[0] = nums[0];
        skip[0] = 0;
        for(int i = 1; i < n; ++i){
            if(nums[i - 1] == nums[i]){
                pick[i] = pick[i - 1] + nums[i];
                skip[i] = skip[i - 1];
            }else if(nums[i - 1] + 1 == nums[i]){
                pick[i] = skip[i - 1] + nums[i];
                skip[i] = max(pick[i - 1], skip[i - 1]);
            }else{
                pick[i] = max(pick[i - 1], skip[i - 1]) + nums[i];
                skip[i] = max(pick[i - 1], skip[i - 1]);
            }
        }
        return max(pick[n - 1], skip[n - 1]);
    }
};

Other solutions

Transform to the House Robber problem.

Reference: [Java/C++] Clean Code with Explanation

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> sums(10001);
        int max_num = 0;
        for(int num: nums){
            sums[num] += num;
            max_num = max(max_num, num);
        }
        int pick = 0, skip = 0;
        for(int i = 1; i <= max_num; ++i){
            int max_pts = max(pick, skip);
            pick = skip + sums[i];
            skip = max_pts;
        }
        return max(pick, skip);
    }
};

Reference: Sharing my Simple Straight Forward Java O(n) Solution -- Explanation Included

class Solution {
public:
    int deleteAndEarn(vector<int>& nums) {
        vector<int> sums(10001);
        int max_num = 0;
        for(int num: nums){
            sums[num] += num;
            max_num = max(max_num, num);
        }
        for(int i = 2; i <= max_num; ++i){
            sums[i] = max(sums[i - 1], sums[i - 2] + sums[i]);
        }
        return sums[max_num];
    }
};