740. Delete and Earn

Medium

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You are given an integer array nums. You want to maximize the number of points you get by performing the following operation any number of times:

• Pick any nums[i] and delete it to earn nums[i] points. Afterwards, you must delete every element equal to nums[i] - 1 and every element equal to nums[i] + 1.

Return the maximum number of points you can earn by applying the above operation some number of times.

Example 1:

Input: nums = [3,4,2]
Output: 6
Explanation: You can perform the following operations:
- Delete 4 to earn 4 points. Consequently, 3 is also deleted. nums = [2].
- Delete 2 to earn 2 points. nums = [].
You earn a total of 6 points.

Example 2:

Input: nums = [2,2,3,3,3,4]
Output: 9
Explanation: You can perform the following operations:
- Delete a 3 to earn 3 points. All 2's and 4's are also deleted. nums = [3,3].
- Delete a 3 again to earn 3 points. nums = [3].
- Delete a 3 once more to earn 3 points. nums = [].
You earn a total of 9 points.

Constraints:

• 1 <= nums.length <= 2 * 104

• 1 <= nums[i] <= 104

class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        d = defaultdict(int)
        # Count Number of Occurences
        for num in nums:
            d[num] += 1
        # If Key 2 Appears 2 Times then its total will be simply key*val
        for key,val in d.items():
            d[key] = key*val
        max_val = max(d.keys())
        l = [0]*(max_val+1)
        for index in range(max_val+1):
            if index in d:
                l[index] = d[index]
        return self.houseRobber(l)
        
    def houseRobber(self, nums) -> int:
        if len(nums)==1:
            return nums[0]
        elif len(nums) == 2:
            return max(nums[0], nums[1])
        nums[2] = nums[2] + nums[0]
        for index in range(3, len(nums)):
            nums[index] = nums[index] + max(nums[index-2], nums[index-3])
        return max(nums[len(nums)-1], nums[len(nums)-2])