<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) {
        sort(digits.begin(), digits.end());
        reverse(digits.begin(), digits.end());
        vector<int> vec[3];
        int tot_sum = 0;
        for(auto &x: digits) {
            vec[ (x%3) ].push_back(x);
            tot_sum += x;
        }

        int val = tot_sum%3;
        if (val!=0){
            if(!vec[val].empty()) {
                vec[val].pop_back();
            } else {
                vec[3-val].pop_back();
                vec[3-val].pop_back();
            }
        }

        
        vector<int> elements;
        for(int i = 0; i<3; i++) {
            for(auto &x:vec[i]){
                elements.push_back(x);
            }
        }
        sort(elements.begin(), elements.end());

        string ans = "";
        while(!elements.empty()){
             if ((ans.size() == 1 )&& ans[0] == '0') {
                ans.pop_back();
            }
            int x = elements.back();
            elements.pop_back();
            ans += ('0' + x);
        }

        return ans;
    }
};
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