<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:
        c = Counter(digits)

        r = sum(digits) % 3
        dic = [[], [1,4,7,2,5,8], [2,5,8,1,4,7]]
        while r:
            for item in dic[r]:
                if c[item] > 0:
                    c[item] -= 1
                    r = (r - item%3)%3
                    break
        ans = []
        for i in range(9, -1,-1):
            ans.extend([str(i)]*c[i])
        if set(ans) == set(["0"]):
            return "0"
        return "".join(ans)
123