<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:
        # define dp[i][k] where 0 <= i < n, and 0 <= k < 2 such that
        # dp[i][k] = s if there exists subsequence of sorteddigits[i : ], which sums to k (mod 3), 
        #            and s is max length of such subsequence
        # dp[i][k] = -1 if there is no such subsequence
        n = len(digits)
        digits.sort(reverse=True)
        dp = [[-1 for _ in range(3)] for _ in range(n + 1)]
        dp[n][0] = 0
        dp[n - 1][0] = 0
        dp[n - 1][digits[n - 1] % 3] = 1

        for i in reversed(range(n - 1)):
            for j in range(3):
                dp[i][j] = dp[i + 1][j]
                temp = (j + 9 - digits[i]) % 3
                if dp[i + 1][temp] != -1:
                    dp[i][j] = max(dp[i][j], 1 + dp[i + 1][temp])
        
        # want to traverse dp[0][0] back
        ans = ''
        idx = 0
        curr_mod = 0
        for i in range(dp[0][0]):
            while idx < n:
                temp = (curr_mod + 9 - digits[idx]) % 3
                if dp[idx + 1][temp] != -1 and dp[0][0] == i + 1 + dp[idx + 1][temp]:
                    ans += str(digits[idx])
                    idx += 1
                    curr_mod = temp
                    break
                else:
                    idx += 1

        if len(ans) > 0 and ans[0] == '0':
            return '0'
            
        return ans

        
255