<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:
        @cache
        def intersect(w1, w2):
            counts = defaultdict(int)
            for letter in w2:
                counts[letter] += 1
            output = ""
            hi = False
            for letter in w1:
                if counts[letter] > 0:
                    counts[letter] -= 1
                    hi= True
                    continue

                output+= letter
            if hi:
                return output
            else:
                return None
            
        
        heap = [(0, target)]
        # Number of nodes is 2^15
        distances = defaultdict(lambda : sys.maxsize)
        seen = set()

        while heap:

            (distance, node) = heapq.heappop(heap)
            
            if (distance, node) in seen:
                continue
            seen.add((distance, node))

            # Each possible edge from here
            for sticker in stickers:
                edge = intersect(node, sticker)
                # We can make progress
                if edge is not None:
                    if distance + 1 < distances[edge]:
                        distances[edge] = distance + 1
                        heapq.heappush(heap, (distance + 1, edge))

        if distances[""] == sys.maxsize:
            return -1 
        return distances[""]


        
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