<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>

INF = float("inf")
class Solution:
    def minStickers(self, stickers: List[str], target: str) -> int:
        # dfs
        self.cache = dict()
        self.stickers = stickers
        self.n = len(stickers)
        target_letter_cnt = dict()
        for c in target:
            if c not in target_letter_cnt:
                target_letter_cnt[c] = 0
            target_letter_cnt[c] += 1
        ans = self.dfs(0, target_letter_cnt)
        return ans if ans != INF else -1

    def dfs(self, idx: int, target_letter_cnt: dict):
        if len(target_letter_cnt) == 0:
            return 0
        if idx == self.n:
            return INF
        cache_key = (idx, str(target_letter_cnt))
        if cache_key in self.cache:
            return self.cache[cache_key]
        # option1: use next sticker
        ans1 = self.dfs(idx+1, dict(target_letter_cnt))

        # option2: use current sticker
        sticker = self.stickers[idx]
        
        is_sticker_useful = False # indicate if this sticker is of any use
        for c in sticker:
            if c not in target_letter_cnt:
                continue
            target_letter_cnt[c] -= 1
            is_sticker_useful = True
            if target_letter_cnt[c] == 0:
                del target_letter_cnt[c]
        if not is_sticker_useful: # if this sticker didn't do shit then return
            return ans1

        ans2 = 1 + self.dfs(idx, dict(target_letter_cnt))
        result = min(ans1, ans2)
        self.cache[cache_key] = result
        return result


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