<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)
        n = len(digits)
        dp = [[-1] * 3 for _ in range(n + 1)]
        dp[0][0] = 0

        for i in range(n):
            for j in range(3):
                if dp[i][j] != -1:
                    k = (j + digits[i]) % 3
                    dp[i + 1][k] = max(dp[i + 1][k], dp[i][j] + 1)
                    dp[i + 1][j] = max(dp[i + 1][j], dp[i][j])

        if dp[n][0] == 0:
            return ""

        res = []
        i, j = n, 0
        while i > 0:
            if dp[i][j] == dp[i - 1][j]:
                i -= 1
            else:
                res.append(str(digits[i - 1]))
                j = (j - digits[i - 1]) % 3
                i -= 1

        res = ''.join(res[::-1])
        return res if res == "" or res[0] != '0' else "0"
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