<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:

        freq = Counter(power)
        # print(freq)
        power = list(sorted(freq.keys()))
        # print(power)
        n = len(power)
        dp = [-1]*n
        def dfs(i):

            if i >= n:
                return 0
            if dp[i] != -1: return dp[i]
            ans = 0
            j = bisect.bisect(power,power[i]+2)
            ans = freq[power[i]] * power[i] + dfs(j)
            dp[i] = max(ans,dfs(i+1))
            return dp[i]
        return dfs(0)

        # for i in range(n-1,-1,-1):
        #     for prev in range(i-1,-1,-1):
        #         if prev == -1 or power[i]- power[prev] >= 3:
        #             dp[i] = max(dp[i],freq[power[i]] * power[i] + dfs(i+1,i))
        #         dp[i] = max(dp[i],dfs(i+1,prev))
        # return dp[0]
        # dp = [0]*(n+1)
        # for i in range(n-1,-1,-1):
        #     for prev in range(i-1,-2,-1):
        #         ans = 0
        #         if prev == -1 or power[i]- power[prev] >= 3:
        #             ans = max(ans,freq[power[i]] * power[i] + dp[i+1])
        #         dp[prev+1] = max(ans,dp[prev+1])
        # return dp[0]




        
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