<p>We are given <code>n</code> different types of <code>stickers</code>. Each sticker has a lowercase English word on it.</p>

<p>You would like to spell out the given string <code>target</code> by cutting individual letters from your collection of stickers and rearranging them. You can use each sticker more than once if you want, and you have infinite quantities of each sticker.</p>

<p>Return <em>the minimum number of stickers that you need to spell out </em><code>target</code>. If the task is impossible, return <code>-1</code>.</p>

<p><strong>Note:</strong> In all test cases, all words were chosen randomly from the <code>1000</code> most common US English words, and <code>target</code> was chosen as a concatenation of two random words.</p>

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

<pre>
<strong>Input:</strong> stickers = [&quot;with&quot;,&quot;example&quot;,&quot;science&quot;], target = &quot;thehat&quot;
<strong>Output:</strong> 3
<strong>Explanation:</strong>
We can use 2 &quot;with&quot; stickers, and 1 &quot;example&quot; sticker.
After cutting and rearrange the letters of those stickers, we can form the target &quot;thehat&quot;.
Also, this is the minimum number of stickers necessary to form the target string.
</pre>

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

<pre>
<strong>Input:</strong> stickers = [&quot;notice&quot;,&quot;possible&quot;], target = &quot;basicbasic&quot;
<strong>Output:</strong> -1
Explanation:
We cannot form the target &quot;basicbasic&quot; from cutting letters from the given stickers.
</pre>

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

<ul>
	<li><code>n == stickers.length</code></li>
	<li><code>1 &lt;= n &lt;= 50</code></li>
	<li><code>1 &lt;= stickers[i].length &lt;= 10</code></li>
	<li><code>1 &lt;= target.length &lt;= 15</code></li>
	<li><code>stickers[i]</code> and <code>target</code> consist of lowercase English letters.</li>
</ul>

class Solution {
public:
    int f(int j, vector<string>& stickers, string target,unordered_map<string,unordered_map<int,int>>& dp) {
        if (target.size() == 0) {
            return 0;
        }
        if (j >= stickers.size()) {
            return INT_MAX - 1; // Use a large value to denote impossibility
        }
        
        string newTarget = target;
        sort(newTarget.begin(),newTarget.end());
        if(dp.find(newTarget)!=dp.end() && dp[newTarget].find(j)!=dp[newTarget].end()){
            return dp[newTarget][j];
        }
        
        // Option 1: Skip this sticker
        int np = f(j + 1, stickers, target,dp);
        
        // Option 2: Use this sticker if possible
        newTarget = target;
        for (char c : stickers[j]) {
            size_t pos = newTarget.find(c);
            if (pos != string::npos) {
                newTarget.erase(pos, 1);
            }
        }
        
        int p=INT_MAX;
        if(newTarget!=target)
            p = 1 + f(j, stickers, newTarget,dp);
        
        sort(target.begin(),target.end());
        
        return dp[target][j]= min(p, np);
    }

    int minStickers(vector<string>& stickers, string target) {
        int n = stickers.size();
        
        unordered_map<string,unordered_map<int,int>>dp;

        int result = f(0, stickers, target,dp);
                
        return result >= INT_MAX - 1 ? -1 : result;
    }
};

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