<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()

        @cache
        def dfs(i: int, j: int) -> str:
            if i < 0:
                return 0 if j == 0 else -inf
            x = dfs(i - 1, (j + digits[i]) % 3) + 1
            y = dfs(i - 1, j)
            return max(x, y)

        ans, j = "", 0
        for i in range(len(digits) - 1, -1, -1):
            if dfs(i, j) == dfs(i - 1, (j + digits[i]) % 3) + 1:
                ans = str(digits[i]) if ans == "0" else f"{ans}{digits[i]}"
                j = (j + digits[i]) % 3
        dfs.cache_clear()
        return ans

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