<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:
    def minStickers(self, stickers: List[str], target: str) -> int:
        freqTarget = Counter(target)
        freqArr = []
        for s in stickers:
            freqArr.append(Counter(s))
        
        dp = {}

        def bfs(freqTarget):
            if not freqTarget:
                return 0
            
            hashed = tuple(sorted(freqTarget.items()))
            if hashed in dp:
                return dp[hashed]

            ans = 100000000
            for f in freqArr:
                if any(char in f for char in freqTarget):
                    ans = min(ans, bfs(freqTarget - f) + 1)
            dp[hashed] = ans
            return ans
        
        result = bfs(freqTarget)
        if result == 100000000:
            return -1
        else:
            return result
            
            

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