33. Delete and Earn

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.

Solution:(DP )

Similar to house robber with little variations We calc the max points by including and excluding a point

S1

class Solution
{
public:
    int deleteAndEarn(vector<int> &nums)
    {

        map<int, int> m;
        for (int i = 0; i < nums.size(); i++)
        {
            m[nums[i]]++;
        }

        vector<int> v;
        for (auto itr = m.begin(); itr != m.end(); itr++)
        {
            v.push_back(itr->first);
        }

        int prev = -1;
        int use = 0;
        int avoid = 0;

        for (int i = 0; i < v.size(); i++)
        {
            int score = 0;
            if (v[i]-1 == prev)
            {
                score = v[i] * m[v[i]] + avoid;
            }
            else
            {
                score = v[i] * m[v[i]] + use;
            }
            avoid = use;
            use = max(use, score);
            prev = v[i];
        }

        return use;
    }
};

S2 (Optimized space)

class Solution
{
public:
    int deleteAndEarn(vector<int> &nums)
    {

        map<int, int> m;
        for (int i = 0; i < nums.size(); i++)
        {
            m[nums[i]]++;
        }

        int prev = -1;
        int use = 0;
        int avoid = 0;

        for (auto itr = m.begin(); itr != m.end(); itr++)
        {
            int score = 0;
            if (itr->first - 1 == prev)
            {
                score = itr->first * itr->second + avoid;
            }
            else
            {
                score = itr->first * itr->second + use;
            }
            avoid = use;
            use = max(use, score);
            prev = itr->first;
        }

        return use;
    }
};