<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, digits: List[int]) -> str:
        SUM = sum(digits)

        remainder_groups = {0: [], 1: [], 2: []}
        for digit in digits:
            remainder_groups[digit % 3].append(digit)
        remainder_groups[1].sort(reverse=True)
        remainder_groups[2].sort(reverse=True)

        if SUM % 3 == 1:
            if len(remainder_groups[1]) >= 1:
                remainder_groups[1].pop()
            elif len(remainder_groups[2]) >= 2:
                remainder_groups[2].pop()
                remainder_groups[2].pop()
        elif SUM % 3 == 2:
            if len(remainder_groups[2]) >= 1:
                remainder_groups[2].pop()
            elif len(remainder_groups[1]) >= 2:
                remainder_groups[1].pop()
                remainder_groups[1].pop()

        result = remainder_groups[0] + remainder_groups[1] + remainder_groups[2]
        SUM = sum(result)
        if SUM % 3 != 0 or len(result) == 0:
            return ""
        result.sort(reverse=True)

        if not result or result[0] == 0:
            return "0"

        return ''.join(map(str, result))
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