<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 typing import List

class Solution:
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        # Sort digits in descending order
        digits.sort(reverse=True)
        
        # Calculate the sum of digits
        total_sum = sum(digits)
        
        # Helper function to remove the smallest digit with a specific modulo
        def remove_smallest_mod(digits, mod):
            for i in range(len(digits) - 1, -1, -1):
                if digits[i] % 3 == mod:
                    return digits[:i] + digits[i+1:]
            return digits
        
        # Adjust digits based on the sum modulo 3
        if total_sum % 3 == 1:
            # Try to remove one digit with mod 1, or two with mod 2
            digits = remove_smallest_mod(digits, 1)
            if sum(digits) % 3 != 0:
                digits = remove_smallest_mod(digits, 2)
                digits = remove_smallest_mod(digits, 2)
        elif total_sum % 3 == 2:
            # Try to remove one digit with mod 2, or two with mod 1
            digits = remove_smallest_mod(digits, 2)
            if sum(digits) % 3 != 0:
                digits = remove_smallest_mod(digits, 1)
                digits = remove_smallest_mod(digits, 1)
        
        # Form the largest number
        result = ''.join(map(str, digits))
        
        # Handle leading zeros
        if result and result[0] == '0':
            return '0'
        
        return result

# Example usage:
# solution = Solution()
# print(solution.largestMultipleOfThree([8, 1, 9]))  # Output: "981"
# print(solution.largestMultipleOfThree([8, 6, 7, 1, 0]))  # Output: "8760"
# print(solution.largestMultipleOfThree([1]))  # Output: ""
17600