<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>

sys.set_int_max_str_digits(10000)
class Solution:
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        c = defaultdict(list)
        for x in digits:
            heappush(c[x%3], x)

        tot = sum(digits)
        
        if tot %3 == 1:
            if 1 in c:
                heappop(c[1])
            else:
                heappop(c[2])
                heappop(c[2])
            
        elif tot % 3 == 2:
            if 2 in c:
                heappop(c[2])
            else:
                heappop(c[1])
                heappop(c[1])
        
        ans = ''
        for v in c.values():
            ans += ''.join(map(str, v))

        ans = ''.join(sorted(ans, reverse=True))
        return str(int(ans)) if ans else ''
142