<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>

from collections import Counter
class Solution:
    def minStickers(self, stickers: List[str], target: str) -> int:
        
        # dfs with memo (exhaustive search)

        memo = {}
        stickers.sort(key = lambda x:len(x),reverse = True)
        # time complexity O(m*2^n)
        def dfs(stickers, target, idx):

            # if no target anymore, just return 0
            if target == "":
                return 0
            
            # if already search through all stickers but still couldn't complete
            if idx == len(stickers):
                return float('inf')

            # characters can be spelled in arbitrary order, so use dict instead of array
            key = (idx, target)
            if key in memo:
                return memo[key]
            
            # choice 1: don't use current sticker
            res = dfs(stickers, target, idx+1)

            # choice 2: use current sticker
            sticker = stickers[idx]
            removed = False

            dic = Counter(sticker)
            curr = []
            for s in target:
                if s in dic and dic[s]>0:
                    dic[s] -= 1
                    if dic[s] == 0:
                        del dic[s]
                    removed = True
                    continue
                else:
                    curr.append(s)
            new_target = "".join(curr)

            # if something can be removed, then do recursion on top of current result
            if removed:
                res = min(res, 1+dfs(stickers, new_target, idx))
            
            memo[key] = res

            return res
        
        res = dfs(stickers, target, 0)

        return res if res != float('inf') else -1
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