<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):
        cnts = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
        for d in digits:cnts[d] += 1
        if (ods := sum(cnts[1:])) == 0:return '0'
        s = sum(i*c for i, c in enumerate(cnts)) % 3
        if s == 0:return ''.join(str(d)*cnts[d] for d in range(9, -1, -1))
        if ods == 1:return '0' if cnts[0] else ''
        prime, sub = s, 1 if s == 2 else 2
        for i in range(prime, 10, 3):
            if cnts[i] > 0:
                cnts[i] -= 1
                return ''.join(str(d)*cnts[d] for d in range(9, -1, -1))
        if ods == 2:return '0' if cnts[0] else ''
        rem = 2
        for i in range(sub, 10, 3):
            if rem == 0 or cnts[i] >= rem:
                if rem != 0: cnts[i] -= rem
                return ''.join(str(d)*cnts[d] for d in range(9, -1, -1))
            if cnts[i] > 0:
                rem -= cnts[i]
                cnts[i] = 0
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