<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:
    int rec(int ind,int sum,map<pair<int,int>,int>&dp,vector<int>&arr){
        if(ind==arr.size()){
            if(sum==0){
                return dp[{ind,sum}]=0;
            }
            return dp[{ind,sum}]=-1e9;
        }
        else if(dp.find({ind,sum})!=dp.end()){
            return dp[{ind,sum}];
        }
        int choosed=1+rec(ind+1,(sum+arr[ind])%3,dp,arr);
        int not_choosed=rec(ind+1,sum%3,dp,arr);
        return dp[{ind,sum}]=max(choosed,not_choosed);
    }
    string largestMultipleOfThree(vector<int>& arr) {
        sort(arr.rbegin(),arr.rend());
        if(arr[0]==0){
            return "0";
        }
        int n=arr.size();
        map<pair<int,int>,int> dp;
        int ans=rec(0,0,dp,arr);
        string s="";
        int sum=0;
        for(int i=0;i<n;i++){
            if(dp[{i+1,sum%3}]>1+dp[{i+1,(sum+arr[i])%3}]){
                continue;
            }
            else{
                s+=('0'+arr[i]);
                sum+=arr[i];
                sum%=3;
            }
        }
        int c=0;
        for(int i=0;i<s.size();i++){
            if(s[i]=='0'){
                c++;
            }
        }
        if(s.size()==c and !s.empty()){
            return "0";
        }
        return s;
    }
};
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