<p>Given an array of integers <code>cost</code> and an integer <code>target</code>, return <em>the <strong>maximum</strong> integer you can paint under the following rules</em>:</p>

<ul>
	<li>The cost of painting a digit <code>(i + 1)</code> is given by <code>cost[i]</code> (<strong>0-indexed</strong>).</li>
	<li>The total cost used must be equal to <code>target</code>.</li>
	<li>The integer does not have <code>0</code> digits.</li>
</ul>

<p>Since the answer may be very large, return it as a string. If there is no way to paint any integer given the condition, return <code>&quot;0&quot;</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> cost = [4,3,2,5,6,7,2,5,5], target = 9
<strong>Output:</strong> &quot;7772&quot;
<strong>Explanation:</strong> The cost to paint the digit &#39;7&#39; is 2, and the digit &#39;2&#39; is 3. Then cost(&quot;7772&quot;) = 2*3+ 3*1 = 9. You could also paint &quot;977&quot;, but &quot;7772&quot; is the largest number.
<strong>Digit    cost</strong>
  1  -&gt;   4
  2  -&gt;   3
  3  -&gt;   2
  4  -&gt;   5
  5  -&gt;   6
  6  -&gt;   7
  7  -&gt;   2
  8  -&gt;   5
  9  -&gt;   5
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> cost = [7,6,5,5,5,6,8,7,8], target = 12
<strong>Output:</strong> &quot;85&quot;
<strong>Explanation:</strong> The cost to paint the digit &#39;8&#39; is 7, and the digit &#39;5&#39; is 5. Then cost(&quot;85&quot;) = 7 + 5 = 12.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> cost = [2,4,6,2,4,6,4,4,4], target = 5
<strong>Output:</strong> &quot;0&quot;
<strong>Explanation:</strong> It is impossible to paint any integer with total cost equal to target.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>cost.length == 9</code></li>
	<li><code>1 &lt;= cost[i], target &lt;= 5000</code></li>
</ul>

class Solution {
public:
//  vector<vector<string>> dp ;
    string MAX(string &a,string &b){
        
        if(a.find("0")!=string::npos) return b;
        if(b.find("0")!=string::npos) return a;
        if(a.size()<b.size())return b;
         
        return a;
    }
// string isans(int idx,int target ,vector<int>& cost){
//     if(idx<0 && target==0)return "";
//     if(idx<0||target<0)return "0";
//     // cout<<idx<<" "<<target<<endl;
//     if(  dp[idx][target]!="-1")return   dp[idx][target];
//     dp[idx][target]=isans(idx-1,target,cost);
//     if(cost[idx]<=target){
//         string temp=(to_string(idx+1)+isans(idx,target-cost[idx],cost));
//         // cout<<temp<<" "<< dp[idx][target]<<" ";
//          dp[idx][target]= MAX(temp,dp[idx][target]);
//          }
//         //  cout<<idx<<target<<dp[idx][target]<<endl;
// return dp[idx][target];

// }
    string largestNumber(vector<int>& cost, int target) {
       // dp.resize(10,vector<string>(target+1,"-1"));
vector<string> dp(target+10,"");
dp[0]=""; dp[target]="0";
        for(int i=1;i<=target;i++){
            for(int j=8;j>=0;j--){
                if(i-cost[j]<0)continue;
                if(dp[i-cost[j]]==""&&i-cost[j]!=0)continue;
               string a=to_string(j+1) + dp[i-cost[j]];
              // cout<<i<<" "<< dp[i]<<" a"<<a<<endl;
               dp[i]=MAX(dp[i],a);
            }
        }
        
        return dp[target];
    }
};
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