<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, d: List[int]) -> str:
        d1 = sorted([i for i in d if i%3 ==1])
        d2 = sorted([i for i in d if i%3 ==2])
        d3 = [i for i in d if i%3 ==0]
        if sum(d) % 3 == 1:
            if len(d1) != 0:
                res = d1[1:] + d2 + d3
            else:
                res = d2[2:]+ d3
        elif sum(d) % 3 == 2:
            if len(d2) != 0:
                res = d1 + d2[1:] + d3
            else:
                res = d1[2:] +d3
        else:
            res = d
        res.sort(reverse = True)
        if not res: return ''
        return str(int(''.join([str(i) for i in res])))
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