<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 solve(int rem,int index,vector<int>&digits,vector<vector<int>>&dp,vector<vector<int>>&dp1){
        if(index==digits.size()){
            if(rem==0){
                return 0;
            }else{
                return -1e9;
            }
        }
        if(dp[index][rem]!=INT_MIN)return dp[index][rem];
        int a=solve(rem,index+1,digits,dp,dp1);
        int b=solve((rem+digits[index])%3,index+1,digits,dp,dp1)+1;
        if(a>b){
            dp1[index][rem]=1;
        }else{
            dp1[index][rem]=2;
        }
        return dp[index][rem]= max(a,b);
    }
    string largestMultipleOfThree(vector<int>& digits) {
        int n= digits.size();
        vector<vector<int>>dp(n+1,vector<int>(3,INT_MIN));
        vector<vector<int>>dp1(n+1,vector<int>(3,0));
        sort(digits.begin(),digits.end(),greater<int>());
        solve(0,0,digits,dp,dp1);
        string ans;
        int i=0;
        int j=0;
        while(i<n){
            if(dp1[i][j]==2){
                ans.push_back('0'+digits[i]);
                j=(j+digits[i])%3;
                i++;
            }else{
                i++;
            }
        }
        sort(ans.begin(),ans.end(),greater<int>());
        if(ans[0]=='0'){
            return "0";
        }
        return ans;
    }
};
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