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

<ul>
	<li>Pick any <code>nums[i]</code> and delete it to earn <code>nums[i]</code> points. Afterwards, you must delete <b>every</b> element equal to <code>nums[i] - 1</code> and <strong>every</strong> element equal to <code>nums[i] + 1</code>.</li>
</ul>

<p>Return <em>the <strong>maximum number of points</strong> you can earn by applying the above operation some number of times</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,2]
<strong>Output:</strong> 6
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [2,2,3,3,3,4]
<strong>Output:</strong> 9
<strong>Explanation:</strong> You can perform the following operations:
- Delete a 3 to earn 3 points. All 2&#39;s and 4&#39;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.</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 2 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>4</sup></code></li>
</ul>

class Solution:
    def deleteAndEarn(self, nums: List[int]) -> int:
        num_sum = defaultdict(lambda: 0)
        for num in nums:
            num_sum[num] += num
        
        extras = {}
        for i, (n, s) in enumerate(sorted(num_sum.items())):
            if i < len(num_sum.items()) - 1 and n + 1 not in num_sum:
                extras[n + 1] = 0
        
        merged = num_sum | extras
        mi = sorted(list(merged.items()))

        if len(mi) == 1:
            return mi[0][1]
        elif len(mi) == 2:
            return max(mi[0][1], mi[1][1])
        elif len(mi) == 3:
            return max(mi[0][1] + mi[2][1], mi[1][1])
        else:
            dp = [0 for _ in range(len(mi))]
            dp[0] = mi[0][1]
            dp[1] = mi[1][1]
            dp[2] = mi[0][1] + mi[2][1]
            for i, (n, s) in enumerate(mi[3:], 3):
                option_1 = s + dp[i - 2]
                option_2 = dp[i - 1] if mi[i - 1][0] + 1 == mi[i][0] else s + dp[i - 1]
                option_3 = s + dp[i - 3]
                dp[i] = max(option_1, option_2, option_3)
        
        return max(dp[-1], dp[-2])

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