<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 {
    unordered_map<int, string> dp;

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

        string maxs = "0";
        for (int i = 0; i < cost.size(); i++) {
            if (target >= cost[i]) {
                string temp = solve(target - cost[i], cost);
                if (temp != "0") {
                    temp = to_string(i + 1) + temp;
                    if (temp.size() > maxs.size() || (temp.size() == maxs.size() && temp > maxs)) {
                        maxs = temp;
                    }
                }
            }
        }
        return dp[target] = maxs;
    }

public:
    string largestNumber(vector<int>& cost, int target) {
        string ans = solve(target, cost);
        return ans == "0" ? "0" : ans;
    }
};

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