<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:
        # t_count = collections.Counter(target)
        # A = [collections.Counter(sticker) & t_count
        #      for sticker in stickers]

        # for i in range(len(A) - 1, -1, -1):
        #     if any(A[i] == A[i] & A[j] for j in range(len(A)) if i != j):
        #         A.pop(i)
        # print("A", A)
        # stickers = ["".join(s_count.elements()) for s_count in A]
        # print("stickers", stickers)
        
        @cache
        def dp(i: int, t: int) -> int:
            if t == (1 << len(target)) - 1: return 0
            if i == len(stickers): return float("inf")
            res = dp(i + 1, t)
            new_t = t
            for l in stickers[i]:
                for j, c in enumerate(target):
                    if new_t & (1 << j): continue
                    if c == l:
                        new_t |= 1 << j
                        break
            if new_t != t: res = min(res, 1 + dp(i, new_t))
            return res


        res = dp(0, 0)
        return res if res != float('inf') else -1
        # return res if res != float("inf") else -1
87340