Largest Multiple of Three
<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>
Greedy; Array; Math; Dynamic Programming; Sorting
class Solution {
    public String largestMultipleOfThree(int[] digits) {
        Arrays.sort(digits);
        int n = digits.length;
        int[][] f = new int[n + 1][3];
        final int inf = 1 << 30;
        for (var g : f) {
            Arrays.fill(g, -inf);
        }
        f[0][0] = 0;
        for (int i = 1; i <= n; ++i) {
            for (int j = 0; j < 3; ++j) {
                f[i][j] = Math.max(f[i - 1][j], f[i - 1][(j - digits[i - 1] % 3 + 3) % 3] + 1);
            }
        }
        if (f[n][0] <= 0) {
            return "";
        }
        StringBuilder sb = new StringBuilder();
        for (int i = n, j = 0; i > 0; --i) {
            int k = (j - digits[i - 1] % 3 + 3) % 3;
            if (f[i - 1][k] + 1 == f[i][j]) {
                sb.append(digits[i - 1]);
                j = k;
            }
        }
        int i = 0;
        while (i < sb.length() - 1 && sb.charAt(i) == '0') {
            ++i;
        }
        return sb.substring(i);
    }
}