<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:
        # Count the frequency of each digit
        count = [0] * 10
        for digit in digits:
            count[digit] += 1
        
        # Calculate the sum of all digits
        total_sum = sum(d * count[d] for d in range(10))
        
        # Function to remove the smallest digit with a specific remainder
        def remove_smallest(rem):
            for i in range(1, 10):
                if i % 3 == rem and count[i] > 0:
                    count[i] -= 1
                    return True
            return False
        
        # Adjust the digits based on the remainder of the sum
        remainder = total_sum % 3
        if remainder == 1:
            if not remove_smallest(1):
                remove_smallest(2)
                remove_smallest(2)
        elif remainder == 2:
            if not remove_smallest(2):
                remove_smallest(1)
                remove_smallest(1)
        
        # Construct the largest number possible
        result = []
        for i in range(9, -1, -1):
            result.extend([str(i)] * count[i])
        
        # Join the result and handle the case with leading zeros
        result_str = ''.join(result)
        if result_str and result_str[0] == '0':
            return '0'
        
        return result_str

# Example usage:
# sol = Solution()
# print(sol.largestMultipleOfThree([8, 1, 9]))  # Output: "981"
# print(sol.largestMultipleOfThree([8, 6, 7, 1, 0]))  # Output: "8760"
# print(sol.largestMultipleOfThree([1]))  # Output: ""
16900