<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:
        memo = {} 
        t = Counter(target)       
        def add(arr, new_counter):
            for i in range(len(arr)):
                old_counter = arr[i]
                merge = old_counter | new_counter
                if merge == new_counter:
                    arr[i] = new_counter.copy()
                    return
                if old_counter | new_counter == old_counter:
                    return
            arr.append(new_counter)
        def helper():
            arr = []
            for sticker in stickers:
                counter = Counter()
                for c in sticker:
                    if c in t:
                        counter[c] += 1
                if counter:
                    add(arr, counter)
            return arr
        stickers = helper()
        def dfs(i, t, cur_len):
            if cur_len <= 0:
                return 0
            if i == len(stickers):
                return float('inf')
            key = (i, t.values(), cur_len)
            if key in memo:
                return memo[key]
            counter = stickers[i]
            memo[key] = dfs(i + 1, t, cur_len)
            new_len = cur_len
            arr = []
            for k, v in counter.items():
                if k in t and t[k] > 0:
                    d = min(t[k], v)
                    arr.append([k, d])
                    t[k] -= d
                    new_len -= d
            if arr:
                # print(i, t, new_len)
                memo[key] = min(memo[key], 1 + dfs(i, t, new_len))
                for k,d in arr:
                    t[k] += d
                    new_len += d
            return memo[key]
        ans = dfs(0, Counter(target), len(target))
        return -1 if ans == float('inf') else ans
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