<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()
    rem = [d % 3 for d in digits]
    s = sum(rem) % 3

    def remove_smallest_with_rem(x): 
      idx = rem.index(x)
      del digits[idx]
      del rem[idx]

    if s != 0:
      try:
        remove_smallest_with_rem(s)
      except ValueError:
        remove_smallest_with_rem(3-s)
        remove_smallest_with_rem(3-s)

    if len(digits) == 0: return "" 
    if digits[-1] == 0: return "0"
    return ("".join(str(d) for d in reversed(digits)))
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