<p>Given an array of digits <code>digits</code>, return <em>the largest multiple of <strong>three</strong> that can be formed by concatenating some of the given digits in <strong>any order</strong></em>. If there is no answer return an empty string.</p>

<p>Since the answer may not fit in an integer data type, return the answer as a string. Note that the returning answer must not contain unnecessary leading zeros.</p>

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

<pre>
<strong>Input:</strong> digits = [8,1,9]
<strong>Output:</strong> &quot;981&quot;
</pre>

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

<pre>
<strong>Input:</strong> digits = [8,6,7,1,0]
<strong>Output:</strong> &quot;8760&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> digits = [1]
<strong>Output:</strong> &quot;&quot;
</pre>

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

<ul>
	<li><code>1 &lt;= digits.length &lt;= 10<sup>4</sup></code></li>
	<li><code>0 &lt;= digits[i] &lt;= 9</code></li>
</ul>

class Solution:
    def del1(self,ans):
        i=len(ans)-1
        ones={"1","4","7"}
        while i>-1 and ans[i] not in ones:i-=1
        if i==-1:return self.del2(self.del2(ans))
        return ans[:i]+ans[i+1:]

    def del2(self,ans):
        i=len(ans)-1
        twos={"2","5","8"}
        while i>-1 and ans[i] not in twos:i-=1
        if i==-1:return self.del1(self.del1(ans))
        return ans[:i]+ans[i+1:]
        

    def largestMultipleOfThree(self, digits: List[int]) -> str:
        c=Counter(digits)
        ans,rem="",0
        for i in range(9,0,-1):
            ans+=str(i)*c[i]
            rem=(rem+i*c[i]%3)%3
        
        if rem==1:ans=self.del1(ans)
        if rem==2:ans=self.del2(ans)
        return ans+"0"*c[0] if ans else "0" if 0 in c else ""
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