<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 check(string a, string b) {
        int l1=a.length(),l2=b.length();
        for(int i=0;i<l2;i++){
            if(b[i]=='-') return a;
        }
        if (l1<l2) {
            return b;
        } else if (l1==l2) {
            return max(a, b);
        } else {
            return a;
        }
    }
    string func(int num, int& n, vector<int>& v1, unordered_map<int, int>& m1,vector<string>&dp) {
        if (num < 0)
            return "-";
        if (num == 0)
            return "";
        if(dp[num]!="a") return dp[num];
        string ans = "-";
        for (int i = 0; i < n; i++) {
            string val;
            char x = m1[v1[i]] + '0';
            val.push_back(x);
            ans = check(ans, val + func(num - v1[i], n, v1, m1,dp));
        }
        return dp[num]=ans;
    }
    string largestNumber(vector<int>& cost, int target) {
        unordered_map<int, int> m1;
        for (int i = 0; i < 9; i++) {
            m1[cost[i]] = max(m1[cost[i]], i + 1);
        }
        vector<int> v1;
        for (auto& x : m1) {
            v1.push_back(x.first);
        }
        int n = v1.size();
        vector<string> dp(target+1,"a");
        string t=func(target, n, v1, m1,dp);
        return t=="-"?"0":t;
    }
};
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