<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:
        # Convert stickers to list of Counters
        sticker_counters = [Counter(sticker) for sticker in stickers]
        target_counter = Counter(target)
        
        # Use memoization to avoid redundant calculations
        result = self.dfs(sticker_counters, target_counter, 0, {})
        return result if result != float("inf") else -1
    
    def dfs(self, stickers, target_counter, idx, memo):
        # Base case: if target is empty, no more stickers are needed
        if not target_counter:
            return 0
        
        # If we've exhausted all stickers and still have a non-empty target, return infinity
        if idx == len(stickers):
            return float("inf")
        
        # Memoization key is the current index and the remaining target as a tuple
        key = (idx, tuple(target_counter.items()))
        if key in memo:
            return memo[key]
        
        # Option 1: Skip the current sticker and try the next one
        result = self.dfs(stickers, target_counter, idx + 1, memo)

        # Option 2: Use the current sticker if it contributes to the target
        current_sticker = stickers[idx]
        if any(c in target_counter for c in current_sticker):
            # Subtract the sticker characters from the target
            new_target_counter = target_counter - current_sticker
            
            # Recursively attempt to solve with the updated target and the same sticker
            result = min(result, 1 + self.dfs(stickers, new_target_counter, idx, memo))

        # Store the result in the memo dictionary
        memo[key] = result
        return result
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