<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 largestMultipleOfThree(self, a: List[int]) -> str:
        
        a.sort(reverse=True)
        
        ones = [x for x in a if x % 3 == 1]
        twos = [x for x in a if x % 3 == 2]
        
        tot = sum(a)
        
        if tot % 3 == 1:
            if ones:
                a.remove(ones[-1])
            else:
                a.remove(twos[-1])
                a.remove(twos[-2])
        elif tot % 3 == 2:
            if twos:
                a.remove(twos[-1])
            else:
                a.remove(ones[-1])
                a.remove(ones[-2])
        
        if not a:
            return ""
        return "".join(map(str,a)).lstrip("0") or "0"
            
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