<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:
        ones, twos, ans, fre = 0, 0, '', [0 for _ in range(10)]

        for i in digits :
            fre[i] += 1 
        
        ones , twos = fre[1] + fre[4] + fre[7], fre[2] + fre[5] + fre[8]
        
        rem = (ones % 3 + 2 * (twos % 3)) % 3

        if rem == 1 :
            if ones > 0:
                ones -= 1 
            elif twos > 1 :
                twos -= 2 
        if rem == 2 :
            if twos > 0 :
                twos -= 1 
            elif ones > 1 :
                ones -= 2
        
        for z, t, o in ((9, 8, 7), (6, 5, 4), (3, 2, 1)) :
            ans += str(z) * fre[z]
            ans += str(t) * min(fre[t], twos)
            twos -= min(fre[t], twos)
            ans += str(o) * min(fre[o], ones)
            ones -= min(fre[o], ones)
        if ans :
            ans += '0' * fre[0]
        elif fre[0] :
            ans += '0'
        return ans 
121