<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)
        dp = ["","",""]
        for i in range(len(digits)):
            d = digits[i]
            newDP = [dp[0],dp[1],dp[2]]
            if dp[(0-d)%3]:
                newDP[0] = max(dp[0],dp[(0-d)%3]+str(d),key=lambda x:(len(x),x))
            if dp[(1-d)%3]:
                newDP[1] = max(dp[1],dp[(1-d)%3]+str(d),key=lambda x:(len(x),x))
            if dp[(2-d)%3]:
                newDP[2] = max(dp[2],dp[(2-d)%3]+str(d),key=lambda x:(len(x),x))
            if not newDP[d%3] and (d != 0 or i == len(digits)-1):
                newDP[d%3] = str(d)
            dp = newDP
        return dp[0]
        
370