<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)
        n = len(digits)
        dp = [[""] * 3 for _ in range(n + 1)]
        # print(len(dp), len(dp[0]))
        for i in range(1, n + 1):
            k = digits[i-1] % 3

            if k == 0:
                dp[i][0] = dp[i-1][0] + str(digits[i-1])
                dp[i][1] = valid(dp[i-1][1], str(digits[i-1]))
                dp[i][2] = valid(dp[i-1][2], str(digits[i-1]))
            elif k == 1:
                dp[i][0] = str_max(valid(dp[i-1][2], str(digits[i-1])), dp[i-1][0])
                dp[i][1] = str_max(dp[i-1][0] + str(digits[i-1]), dp[i-1][1])
                dp[i][2] = str_max(valid(dp[i-1][1], str(digits[i-1])), dp[i-1][2])
            elif k == 2:
                dp[i][0] = str_max(valid(dp[i-1][1], str(digits[i-1])), dp[i-1][0])
                dp[i][1] = str_max(valid(dp[i-1][2], str(digits[i-1])), dp[i-1][1])
                dp[i][2] = str_max(dp[i-1][0] + str(digits[i-1]), dp[i-1][2])
        # print(dp)
        if len(dp[n][0]) > 0 and dp[n][0][0] == '0':
            return '0'
        return dp[n][0]

def valid(a, b):
    if a == "":
        return ""
    else:
        return a + b

def str_max(a, b):
    if len(a) > len(b):
        return a
    elif len(a) < len(b):
        return b
    else:
        return max(a, b)

'''
8 7 6 1 0

0 0 87  
1 0  7
2 8  8

'''

388