<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:
        n = len(stickers)
        freqs = []
        for sticker in stickers:
            freq = defaultdict(int)
            for c in sticker:
                freq[c] += 1
            freqs.append(freq)
        
        target_freq = defaultdict(int)
        for c in target:
            target_freq[c] += 1
        
        memo = {}
        
        def dfs(freq):
            if not freq:
                return 0
            
            key = tuple(sorted(freq.items()))
            if key in memo:
                return memo[key]
            
            used = float('inf')
            for s in freqs:
                if any(c in s for c in freq):  # Check if sticker has any needed letter
                    next_freq = freq.copy()
                    progress = False
                    for letter, value in s.items():
                        if letter in next_freq:
                            update = min(value, next_freq[letter])
                            next_freq[letter] -= update
                            if next_freq[letter] == 0:
                                del next_freq[letter]
                            if update > 0:
                                progress = True
                    
                    if progress:  # Only recurse if we've made progress
                        res = dfs(next_freq)
                        if res != -1:
                            used = min(used, 1 + res)
            
            result = used if used != float('inf') else -1
            memo[key] = result
            return result
        
        return dfs(target_freq)
23185