<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:
        digitCount = Counter(digits)
        mod = sum(digits) % 3

        # Trying to remove one digit that has a specific remainder when divided by 3
        def tryRemove(_mod):
            numsPossibleToRemove = [1, 4, 7] if _mod == 1 else [2, 5, 8]
            for num in numsPossibleToRemove:
                if digitCount[num] > 0:
                    digitCount[num] -= 1
                    return True
            return False
        
        # Repeat removing digits until the sum is divisible by 3
        while mod > 0:
            if tryRemove(mod):
                break
            if tryRemove(3 - mod):
                mod = 3 - mod
            else:
                return ''

        # Build the result string by adding digits from largest to smallest
        ans = ''
        for i in range(9, 0, -1):
            if digitCount[i] > 0:
                ans = ans + (str(i) * digitCount[i])

        # Handle edge case when only '0' digits are present
        if ans == '' and 0 in digitCount:
            return '0'
        
        return ans + '0' * digitCount[0]
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