<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:
        l1 = [i for i in digits if i%3 == 1]
        l2 = [i for i in digits if i%3 == 2]
        l3 = [i for i in digits if i%3 == 0]
        
        l1.sort()
        l2.sort()
        l3.sort()
        
        res = []
        if sum(digits) % 3 == 1:
            if l1:
                res = l1[1:] + l2 + l3
            else:
                res = l2[2:] + l3
        elif sum(digits) % 3 == 2:
            if l2:
                res = l1 + l2[1:] + l3
            else:
                res = l1[2:] + l3
        else:
            res = digits

        if not res:
            return ''
        
        res.sort(reverse=True)
        res = [str(i) for i in res]
        
        for i in range(len(res)):
            if res[i] == '0':
                res[i] = ''
            else:
                break
                
        newres = ''.join(res)
        
        return newres if newres else '0'
        
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