<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:
        # Step 1: Sort digits in descending order
        digits.sort(reverse=True)
        
        # Step 2: Calculate the total sum of digits
        total_sum = sum(digits)
        
        # Step 3: Initialize lists to keep track of digits with different remainders
        remainder1_digits = []
        remainder2_digits = []
        
        for d in digits:
            if d % 3 == 1:
                remainder1_digits.append(d)
            elif d % 3 == 2:
                remainder2_digits.append(d)
        
        # Step 4: Handle cases based on remainder of total sum
        if total_sum % 3 == 1:
            # Remove one smallest digit with remainder 1 or two smallest digits with remainder 2
            if remainder1_digits:
                digits.remove(remainder1_digits[-1])
            else:
                digits.remove(remainder2_digits[-1])
                digits.remove(remainder2_digits[-2])
        elif total_sum % 3 == 2:
            # Remove one smallest digit with remainder 2 or two smallest digits with remainder 1
            if remainder2_digits:
                digits.remove(remainder2_digits[-1])
            else:
                digits.remove(remainder1_digits[-1])
                digits.remove(remainder1_digits[-2])
            
        print(digits)
        if len(digits) > 1 and sum(digits) == 0:
            return "0"
        # Step 5: Check for all zeros and convert to a string
        if not digits:
            return ""
        
        if digits[0] == 0:
            return "0"
        
        

        return "".join(map(str, digits))
            
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