<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:
        from collections import defaultdict
        n = len(digits)
        digits = sorted(digits)
        tab = defaultdict(list)
        for i in range(n):
            tab[int(digits[i]) % 3].append(digits[i])
        ans = []
        i, j = len(tab[1]), len(tab[2])
        if i >= 3:
            ans.extend(tab[1][i%3:])
            tab[1] = tab[1][:i%3]
        if j >= 3:
            ans.extend(tab[2][j%3:])
            tab[2] = tab[2][:j%3]
        while tab[1] and tab[2]:
            if 0 < len(tab[1]) < 3 and 0 < len(tab[2]) < 3:
                    x = tab[1].pop()
                    y = tab[2].pop()
                    ans.extend([x, y])
        if len(tab[2]) == 2 and len(ans) % 3 == 0 and ans:
            for i in range(len(ans)):
                if ans[i] % 3 == 1:
                    ans.pop(i)
                    break
            if len(ans) % 3 == 2:
                ans.extend(tab[2])
        if len(tab[1]) == 2 and len(ans) % 3 == 0 and ans:
            for i in range(len(ans)):
                if ans[i] % 3 == 2:
                    ans.pop(i)
                    break
            if len(ans) % 3 == 2:
                ans.extend(tab[1])
        for x in tab[0]:
            ans.append(x)
        if set(ans) == {0}:
            return "0"
        ans = sorted(ans, key=lambda x: -x)
        out = ""
        for x in ans:
            out += str(x)
        return out
147