Delete and earn

Time: O(N); Space: O(1); medium

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:

  • The length of nums is at most 20000.

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

Hints:

  1. If you take a number, you might as well take them all. Keep track of what the value is of the subset of the input with maximum M when you either take or don’t take M.

[1]:

class Solution1(object):
    """
    Time: O(N)
    Space: O(1)
    """
    def deleteAndEarn(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        vals = [0] * 10001
        for num in nums:
            vals[num] += num
        val_i, val_i_1 = vals[0], 0
        for i in range(1, len(vals)):
            val_i_1, val_i_2 = val_i, val_i_1
            val_i = max(vals[i] + val_i_2, val_i_1)

        return val_i

[2]:

s = Solution1()

nums = [3, 4, 2]
assert s.deleteAndEarn(nums) == 6

nums = [2, 2, 3, 3, 3, 4]
assert s.deleteAndEarn(nums) == 9