<p>A magician has various spells.</p>

<p>You are given an array <code>power</code>, where each element represents the damage of a spell. Multiple spells can have the same damage value.</p>

<p>It is a known fact that if a magician decides to cast a spell with a damage of <code>power[i]</code>, they <strong>cannot</strong> cast any spell with a damage of <code>power[i] - 2</code>, <code>power[i] - 1</code>, <code>power[i] + 1</code>, or <code>power[i] + 2</code>.</p>

<p>Each spell can be cast <strong>only once</strong>.</p>

<p>Return the <strong>maximum</strong> possible <em>total damage</em> that a magician can cast.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">power = [1,1,3,4]</span></p>

<p><strong>Output:</strong> <span class="example-io">6</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum possible damage of 6 is produced by casting spells 0, 1, 3 with damage 1, 1, 4.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">power = [7,1,6,6]</span></p>

<p><strong>Output:</strong> <span class="example-io">13</span></p>

<p><strong>Explanation:</strong></p>

<p>The maximum possible damage of 13 is produced by casting spells 1, 2, 3 with damage 1, 6, 6.</p>
</div>

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

<ul>
	<li><code>1 &lt;= power.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= power[i] &lt;= 10<sup>9</sup></code></li>
</ul>

class Solution:
    def maximumTotalDamage(self, power: List[int]) -> int:
        n = len(power)
        # vispower = [False] * (1000001)
        power.sort()
        dc = [[power[0], power[0]]]
        

        for i in range(1, n):
            x = power[i]
            if (x != power[i-1]):
                dc.append([x, x])
            else:
                dc[-1][1] += x

        dp = [-1] * len(dc)
        n = len(dc)

        def dfs(idx):
            if idx >= n:
                return 0
            if dp[idx] != -1:
                return dp[idx]
            ans1, ans2 = 0, 0
            # no take
            ans1 = dfs(idx+1)

            # take
            j = idx+1
            x = dc[idx][0]
            while j<n and dc[j][0] <= x+2:
                j+=1
            ans2 = dc[idx][1] + dfs(j)
            
            # print(idx, ans1, ans2)
            dp[idx] = max(ans1, ans2)
            return dp[idx]
        
        return dfs(0)


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