<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:
        total = sum(digits)
        count = [0] * 10
        for d in digits:
            count[d] += 1

        def remove(digit):
            if count[digit]:
                count[digit] -= 1
                return True
            return False

        if total % 3 == 1:
            if not (remove(1) or remove(4) or remove(7)):
                remove(2) or remove(5) or remove(8)
                remove(2) or remove(5) or remove(8)
        elif total % 3 == 2:
            if not (remove(2) or remove(5) or remove(8)):
                remove(1) or remove(4) or remove(7)
                remove(1) or remove(4) or remove(7)
        ans = "".join(str(i) * count[i] for i in range(9, -1, -1))

        return "0" if ans.startswith("0") else ans or ""
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