<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:
        cnt = Counter(power)  # Count the occurrences of each spell's damage
        power = list(set(power))  # Remove duplicates
        power.sort()  # Sort the power list to allow binary searching
        n = len(power)
        
        # Initialize dp list with -1 to indicate uncalculated subproblems
        dp = [-1] * n
        
        def recur(ind):
            # Base case: If we've gone past the last spell, return 0
            if ind >= n: 
                return 0
            
            # If this subproblem has already been calculated, return the stored value
            if dp[ind] != -1:
                return dp[ind]
            
            # Option 1: Skip the current spell
            notTake = recur(ind + 1)
            
            # Option 2: Take the current spell and skip to the next valid spell
            take = float('-inf')
            pos = bisect_left(power, power[ind] + 2 + 1)  # Find next valid spell index
            take = (cnt[power[ind]] * power[ind]) + recur(pos)
            
            # Store the result in dp array
            dp[ind] = max(take, notTake)
            
            return dp[ind]
    
        return recur(0)

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