<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:
        count = Counter(digits)
        total_sum = sum(digits)
        def remove_digits(rem, count):
            for i in range(1, 10):
                if i % 3 == rem and count[i] > 0:
                    count[i] -= 1
                    return True
            return False
        remainder = total_sum % 3
        if remainder == 1:
            if not remove_digits(1, count):
                remove_digits(2, count)
                remove_digits(2, count)
        elif remainder == 2:
            if not remove_digits(2, count):
                remove_digits(1, count)
                remove_digits(1, count)
        
        result = []
        for i in range(9, -1, -1):
            result.extend([str(i)] * count[i])
        
        if result and result[0] == '0':
            return '0'
        
        return ''.join(result) if result else ''

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