<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>

"""

https://leetcode.com/problems/stickers-to-spell-word/solutions/4095051/python-accepted-dfs-memo-solution-with-explanation-easy-to-understand/?envType=company&envId=tiktok&favoriteSlug=tiktok-three-months&status=ATTEMPTED%2CTO_DO


DP solution 
    1. take and or not take sticker [infinite tickets]
    2. count the number of tickets used 
    3. keep on removing the letter from the target

"""

class Solution:
    def minStickers(self, stickers: List[str], target: str) -> int:
        

        # sticker_lookup = {i : Counter(stickers[i]) for i in range(len(stickers))}
        # target = Counter(target)
        lookup = {}

        def recursion(i, target):
            
            if (i, target) in lookup:
                return lookup[(i, target)]
                
            if i == len(stickers):

                if len(target) == 0:
                    return 0

                return math.inf


            if len(target) == 0:
                return 0
            
            # original = target.copy()
            changed = False

            sticker_lookup = Counter(stickers[i])
            new_string = ""

            for c in target:

                if c in sticker_lookup and sticker_lookup[c] > 0:
                    sticker_lookup[c] -= 1
                    changed = True
                else:
                    new_string = new_string + c
            

                        

            
            take = math.inf
            if changed:
                take = 1 + recursion(i, new_string)
            
            no_take = recursion(i + 1, target)


            lookup[(i, target)] =  min(take, no_take)
            return lookup[(i, target)]

        ans =  recursion(0, target)

        if ans == math.inf:
            return -1

        return ans



            
57730