<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, nums: List[int]) -> str:
        nums.sort()
        n=len(nums)   
        @cache
        def dfs(i,j):
            if i==-1:
                return 0 if j==0 else -inf
            return max(dfs(i-1,j),dfs(i-1,(j-nums[i])%3)+1)
        res = []
        MOD = 0
        SIZE = dfs(n-1,MOD)
        for i in range(n-1,-1,-1):
            if dfs(i,MOD) == dfs(i-1,(MOD-nums[i])%3)+1 and dfs(i,MOD)==SIZE:
                res.append(str(nums[i]))
                SIZE-=1
                MOD = (MOD-nums[i])%3
        val = ''.join(res)
        if not val:
            return ''
        if all(ch=='0' for ch in val):
            return '0'
        return val

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