<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 largestNumber(vector<int>& cost, int target) {
        //dp[i] stores the largest digit that can be fromed with i as target
        vector<string> dp(target+1,"#");
        
        int len=cost.size();
        helper(cost,len,target,dp);
        return dp[target];
        
    }
    string helper(vector<int>& cost,int& len,int target,vector<string>& dp)
    {
        if(target==0)
            return "";
        //the current sub problem is already calculated
        if(dp[target]!="#")
            return dp[target];
        
        //not yet calculated
        dp[target]="0";
        string curr="";
        for(int i=len-1;i>=0;i--)
        {
            //use i if the cost is less than or equal to the target
            if(cost[i]<=target){
                curr.push_back(i+1+'0');
                string x=helper(cost,len,target-cost[i],dp);
                //if the return string is not zero 
                //it means we can build some largest number with the remaining target
                if(x!="0")
                {
                    
                    //we can form some number with the remaining target
                    curr.append(x);
                    int currLen=curr.length();
                    int tLen=dp[target].length();
                    //update if we got the largest number than dp[target]
                    if(tLen < currLen)
                        dp[target]=curr;
                    else if(tLen == currLen)
                        dp[target]=max(dp[target],curr);
                }
                curr="";
            }
                
        }
        //now dp[target] would be either 0 or some digit(if there is some answer) 
        return dp[target];
        
    }
};
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