<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:
    Solution(){
        ios_base::sync_with_stdio(0);
        cin.tie(0);
        cout.tie(0);
    }
    string largestMultipleOfThree(vector<int>& digits) {
       int sum=0;
       vector<vector<int>> r(3);
       sort(digits.rbegin(),digits.rend());
       for(int& d:digits){
           r[d%3].push_back(d);
           sum+=d;
       }
       if(sum%3==1){
        if(r[1].size()>0) r[1].pop_back();
        else{
            r[2].pop_back();
            r[2].pop_back();
        }
       }
       else if(sum%3){
        if(r[2].size()>0) r[2].pop_back();
        else{
            r[1].pop_back();
            r[1].pop_back();
        }
       }
       string ans="";
       priority_queue<int> pq;
       for(int i=0;i<3;i++){
        for(int& e:r[i]) pq.push(e);
       }
       while(!pq.empty()){
        ans.push_back('0'+pq.top()); pq.pop();
       }
       if(ans.size() && ans.front()=='0' && ans.back()=='0') return "0";
       return ans;
    }
};
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