<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)
        INF = 10**20
        
        @cache
        def helper(i, s):
            if i == len(digits):
                if s != 0:
                    return -INF
                return 0
            a = 1 + helper(i + 1, (s + digits[i]) % 3)
            b = helper(i + 1, s)
            return max(a, b)
        
        size = helper(0, 0)
        
        if size <= 0:
            return ""
        
        ans = []
        s = 0
        for i in range(len(digits)):
            if size <= 0:
                break
            if helper(i + 1, (s + digits[i]) % 3) == size - 1:
                ans.append(digits[i])
                s = (s + digits[i]) % 3
                size -= 1
        
        if ans and ans[0] == 0:
            return "0"
        
        return "".join(map(str, ans))
329