<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:
    string dp[5001];  
    string solve(int target,vector<int>& cost){
        if(target ==0) return "";
        if(!dp[target].empty()) return dp[target];

        string maxNum="0";
        for(int i=0;i<9;i++){
            if(target>=cost[i]){
                string result = solve(target-cost[i],cost);
                if(result!="0"){
                    result = to_string(i+1) +result;
                    if(result.length()>maxNum.length() || (result.length() == maxNum.length() && result > maxNum)){
                        maxNum=result;
                    }
                }
            }
        }
        return dp[target]= maxNum;
    }
    string largestNumber(vector<int>& cost, int target) {
        fill(begin(dp), end(dp), "");
        return solve(target, cost) == "0" ? "0" : solve(target, cost);
    }
};
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