<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:
        digits.sort()
        
        mp = defaultdict(list)
        s = sum(digits)%3
        for i, d in enumerate(digits):
            mp[d%3].append(i)
        
        rm = set()
        if s == 1:
            if mp[1]:
                rm.add(mp[1][0])
            elif len(mp[2]) >= 2:
                rm.add(mp[2][0])
                rm.add(mp[2][1])
            else:
                return ""
        if s == 2:
            if mp[2]:
                rm.add(mp[2][0])
            elif len(mp[1]) >= 2:
                rm.add(mp[1][0])
                rm.add(mp[1][1])
            else:
                return ""
        res = []
        for i in range(len(digits)-1,-1,-1):
            if i not in rm:
                res.append(str(digits[i]))
        if len(res) > 1 and res[0] == "0": return "0"
        return "".join(res)
139