<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 __init__(self):
        self.memo={}
    def rec(self, nums, idx, curr):
        if idx>=len(nums):
            if curr==0:
                return "~"
            return ""
        if (idx,curr) in self.memo:
            return self.memo[(idx,curr)]
        takes = str(nums[idx]) + self.rec(nums , idx+1, (curr+nums[idx])%3)
        nontakes = self.rec(nums , idx+1, curr)
        temp = ""
        if "~" in takes:
            temp = takes
        if "~" in nontakes:
            temp = max(takes, nontakes, key = lambda x: (len(x), x))
        self.memo[((idx,curr))] = temp
        return self.memo[((idx,curr))]
    def largestMultipleOfThree(self, digits: List[int]) -> str:
        digits.sort(reverse=True)
        store = self.rec(digits, 0, 0)[:-1]
        if not store:
            return ""
        if store.count("0") == len(store):
            return "0"
        return store
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