<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(reverse=True)
        total = sum(digits)
        mod = total % 3
        if mod == 1:
            # find one digit mod 3 = 1
            for idx in range(len(digits)-1, -1,-1):
                if digits[idx] % 3 == 1:
                    tmp = digits[:idx] + digits[idx+1:]
                    if len(tmp) > 0:
                        return ''.join([str(item) for item in tmp]).lstrip('0') or "0"
            data = []
            # find two digits mod 3 = 2
            for idx in range(len(digits)-1,-1,-1):
                if digits[idx] % 3 == 2:
                    data.append(idx)
                    if len(data) == 2:
                        tmp = digits[:data[1]] + digits[data[1]+1:data[0]] + digits[data[0]+1:]
                        if len(tmp) > 0:
                            return ''.join([str(item) for item in tmp]).lstrip('0') or "0"
            return ""
        elif mod == 2:
            # find one digit mod 3 = 2
            for idx in range(len(digits)-1, -1,-1):
                if digits[idx] % 3 == 2:
                    tmp = digits[:idx] + digits[idx+1:]
                    if len(tmp) > 0:
                        return ''.join([str(item) for item in tmp]).lstrip('0') or "0"
            data = []
            # find two digits mod 3 = 1
            for idx in range(len(digits)-1,-1,-1):
                if digits[idx] % 3 == 1:
                    data.append(idx)
                    if len(data) == 2:
                        tmp = digits[:data[1]] + digits[data[1]+1:data[0]] + digits[data[0]+1:]
                        if len(tmp) > 0:
                            return ''.join([str(item) for item in tmp]).lstrip('0') or "0"
            return ""
        else:
            return ''.join([str(item) for item in digits]).lstrip('0') or "0"

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