<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());
        // int n = digits.size();
        // for(int i = 0; i < n; i++){
        //     int s += n*(n+1) / 2;
        //     if(s % 3 == 0){
        //         v.push_back(s);
        //     }
        //     else
                
        // }
        // return "";
        string res = "";
        int m1[] = {1, 4, 7, 2, 5, 8}, m2[] = {2, 5, 8, 1, 4, 7};
  int sum = 0, ds[10] = {};  
  for (auto d : digits) {
      ++ds[d];
      sum += d;
  }
  while (sum % 3 != 0) {
    for (auto i : sum % 3 == 1 ? m1 : m2) {
      if (ds[i]) {
        --ds[i];
        sum -= i;
        break;
      }
    }
  }
  for (int i = 9; i >= 0; --i)
    res += string(ds[i], '0' + i);
  return res.size() && res[0] == '0' ? "0" : res;
    }
};
23860