<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 {
public:
    string largestMultipleOfThree(vector<int>& digits) {
        string answer = "";
        int carry = 0;

        sort(digits.begin(), digits.end(), greater<int>());
        for(size_t i = 0; i < digits.size(); ++i) {
            carry = (carry + digits[i]) % 3;
            answer += to_string(digits[i]);
        }

        bool ok = false;
        if (carry == 0) {
            ok = true;
        } else if (carry == 1) {
            ok = eraseN(answer, "147", 1);
            
            if (!ok) {
                ok = eraseN(answer, "258", 2);
            }
        } else {
            ok = eraseN(answer, "258", 1);

            if (!ok) {
                ok = eraseN(answer, "147", 2);
            }
        }
        
        bool haveZeros = false;
        bool onlyZeros = true;
        for(size_t i = 0; i < answer.length(); ++i) {
            if (answer[i] != '0') {
                onlyZeros = false;
            }
            if (answer[i] == '0') {
                haveZeros = true;
            }
        }

        return (ok && !onlyZeros) ? answer : haveZeros ? "0" : "";
    }
private:
    bool eraseN(string& s, string const& chars, size_t amount) {
        string original(s);
        for(size_t i = 0; i < amount; ++i) {
            size_t pos = s.find_last_of(chars);
            if (pos != string::npos) {
                s.erase(pos, 1);
            } else {
                s = original;
                return false;
            }
        }

        return true;
    }
};
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