<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(i, 0) = max(1 + dp(i + 1, nums[i] % 3), dp(i, 0))
        # dp(i, 1) = dp(i + 1, )


        dp = []
        for _ in range(len(digits) + 1):
            dp.append([(float('-inf'), None)]  * 3)

        dp[-1][0] = (0, None, None) # curr, next

        for i in reversed(range(len(digits))):
            # dp[i][0]
            for j in range(3):
                use = dp[i + 1][(j + int(digits[i])) % 3]
                dont_use = dp[i + 1][j]
                if 1 + use[0] >= dont_use[0]:
                    dp[i][j] = (1 + use[0], i, use[1])
                else:
                    dp[i][j] = dont_use

        res_lst = []
        curr = dp[i][0]
        tot = 0
        while curr is not None and curr[1] is not None:
            res_lst.append(str(digits[curr[1]]))
            nxt = curr[2]
            tot = (tot + int(digits[curr[1]])) % 3
            curr = dp[nxt][tot] if nxt is not None else None
        res = "".join(res_lst)
        if res and res[0] == "0":
            return "0"
        return res
        


        
        
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