<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>&digits){
        if(ind==digits.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 p=1+rec(ind+1,(sum+digits[ind])%3,dp,digits);
        int np=rec(ind+1,sum%3,dp,digits);
        return dp[{ind,sum}]=max(p,np);
    }
    string largestMultipleOfThree(vector<int>& digits) {
        sort(digits.begin(),digits.end());
        reverse(digits.begin(),digits.end());
        if(digits[0]==0){
            return "0";
        }
        int n=digits.size();
        // vector<vector<int>> dp(n+1,vector<int>(9*n+1,-1));
        map<pair<int,int>,int> dp;
        int ans=rec(0,0,dp,digits);
        string s="";
        int sum=0;
        for(int i=0;i<n;i++){
            if(dp[{i+1,sum%3}]>1+dp[{i+1,(sum+digits[i])%3}]){
                continue;
            }
            else{
                s+=('0'+digits[i]);
                sum+=digits[i];
                sum%=3;
            }
        }
        int cnt=0;
        for(int i=0;i<s.size();i++){
            if(s[i]=='0'){
                cnt++;
            }
        }
        if(s.size()==cnt and !s.empty()){
            return "0";
        }
        return s;
    }
};
45460