<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>

from sortedcontainers import SortedList
class Solution:
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        a=defaultdict(lambda:SortedList([]))
        d=SortedList(digits)
        for i in digits:
            a[i%3].add(i)
        s=sum(d)
        # print(s%3)
        if(s%3==0):
            if(d and d[-1]==0):
                return '0'
            return ''.join([str(i) for i in d[::-1]])
        elif(s%3==2):
            if(a[2]):
                d.discard(a[2][0])
                a[2].discard(a[2][0])
                if(d and d[-1]==0):
                    return '0'
                return ''.join([str(i) for i in d[::-1]])
            else:
                if(len(a[1])>1):
                    d.discard(a[1][0])
                    a[1].discard(a[1][0])
                    d.discard(a[1][0])
                    a[1].discard(a[1][0])
                    if(d and d[-1]==0):
                        return '0'
                    return ''.join([str(i) for i in d[::-1]])
        else:
            print(a[1],a[2])
            if(a[1]):
                d.discard(a[1][0])
                a[1].discard(a[1][0])
                if(d and d[-1]==0):
                    return '0'
                return ''.join([str(i) for i in d[::-1]])
            else:
                if(len(a[2])>1):
                    d.discard(a[2][0])
                    a[2].discard(a[2][0])
                    d.discard(a[2][0])
                    a[2].discard(a[2][0])
                    if(d and d[-1]==0):
                        return '0'
                    return ''.join([str(i) for i in d[::-1]])
        return ''
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