<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>

import sys 
class Solution:
    def largestMultipleOfThree(self, digits):
        sys.set_int_max_str_digits(10000)
        running_sum, ans0, ans1, ans2 = 0, [], [], []
        for i in digits:
            running_sum += i 

            if i%3 == 0:
                ans0.append(i)
            elif i%3 == 1:
                ans1.append(i)
            else:
                ans2.append(i)
        ans1.sort(reverse = True)
        ans2.sort(reverse = True)
        if running_sum%3 == 1:
            if ans1:
                ans1.pop()
            else:
                if ans2:
                    ans2.pop()
                else:
                    return ""
                if ans2:
                    ans2.pop()
                else:
                    return ""
        if running_sum%3 == 2:
            if ans2:
                ans2.pop()
            else:
                if ans1:
                    ans1.pop()
                else:
                    return ""
                if ans1:
                    ans1.pop()
                else:
                    return ""
        result = ans0 + ans1 + ans2
        result.sort(reverse = True)
        s = ""
        for i in result:
            s += str(i)
        return str(int(s)) if result else ""
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