<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>

from collections import Counter

class Solution:
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        remain = ((1, 4, 7, 2, 5, 8), (2, 5, 8, 1, 4, 7))
        total = sum(digits)
        counter = Counter(digits)
        res = []
        
        while total % 3 != 0:
            for i in remain[total % 3 == 2]:
                if counter[i] != 0:
                    total -= i
                    counter[i] -= 1
                    break
                    
        for i in range(9, -1, -1):
            if counter[i]:
                res.extend(str(i) * counter[i])
                
        return '0' if res and res[0] == '0' else ''.join(res)
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