<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:
        digits.sort(reverse=True)
        n = len(digits)
        
        # Calculate the sum of all digits
        total_sum = sum(digits)
        
        # If the total sum is not divisible by 3, remove the smallest digits that make it divisible by 3
        if total_sum % 3 != 0:
            remainder = total_sum % 3
            found = False
            
            # Try to find a single digit that has the same remainder
            for i in range(n - 1, -1, -1):
                if digits[i] % 3 == remainder:
                    digits.pop(i)
                    found = True
                    break
            
            # If no single digit is found, try to find two digits whose sum has the same remainder
            if not found:
                for i in range(n - 1, -1, -1):
                    if not found:
                        for j in range(i - 1, -1, -1):
                            if (digits[i] + digits[j]) % 3 == remainder:
                                digits.pop(i)
                                digits.pop(j)
                                found = True
                                break
            
            # If still not found, return an empty string
            if not found:
                return ""
        
        # Join the remaining digits to form the largest multiple of three
        result = ''.join(map(str, digits))
        
        # Remove leading zeros
        if result.startswith('0'):
            return '0'
        
        return result
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