<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>& nums) {
        string ans="";
        int n=nums.size(),sum=0;
        sort(nums.begin(),nums.end(),greater<int>());
        vector<int> rem0,rem1,rem2;
        for(auto i:nums){
            sum+=i;
            if(i%3==0){
                rem0.push_back(i);
            }
            else if(i%3==1){
                rem1.push_back(i);
            }
            else{
                rem2.push_back(i);
            }
        }
        
        if(sum%3==1){
            if(rem1.size()>0) rem1.pop_back();
            else{
                rem2.pop_back();
                rem2.pop_back();
            }
        }
        else if(sum%3==2){
            if(rem2.size()>0){
                rem2.pop_back();
            }
            else{
                rem1.pop_back();
                rem1.pop_back();
            }
        }
        priority_queue<int> pq;
        for(int i:rem1){
            pq.push(i);
        }
        for(int i:rem2){
            pq.push(i);
        }
        
        if(pq.empty() && rem0.size()>1 && rem0[0]==0){
            pq.push(0);
        }
        else{
            for(int i:rem0){
                pq.push(i);
            }
        }
        while(!pq.empty()){
            int x=pq.top();
            pq.pop();
            string t=to_string(x);
            ans+=t;
        }
        return ans;
    }
};
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