<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 {
public:
    vector<vector<int>> stickerLetterCounts;
    int minStickers(vector<string>& stickers, string target) {
        stickerLetterCounts = vector<vector<int>>(stickers.size(), vector<int>(27, 0)); 
        map<pair<int,string>, long> memo;
        long res = dfs(stickers, target, 0, memo);
        return res >= INT_MAX ? -1 : res;
    }
    // dfs function always return minStickers for target
    long dfs(vector<string>& stickers, string target, int idx, map<pair<int,string>, long>& memo){
        //Successfully find an answer
        if(target == "") 
            return 0;

        //Failed find any answers
        if(idx == stickers.size())
            return INT_MAX;

        //Answer already exist
        pair<int,string> key = {idx, target};
        if(memo.find(key)!=memo.end()){
            return memo[{idx, target}];
        }

        //Keep finding answers
        //choice 1: keep using stickers[idx]

        string newTarget;
        vector<int> currentStickerCount = getCurrentStickerCount(stickers, idx);
        
        for(char c: target){
            if(currentStickerCount[c-'a']>0){
                currentStickerCount[c-'a']--;
            }
            else{
                newTarget += c;
            }
        }

        //choice 2: Skip stickers[idx], use stickers[idx+1]
        long res = dfs(stickers, target, idx+1, memo);


        if(newTarget!=target){
            res = min(res, long(1+ dfs(stickers, newTarget, idx, memo)));
        }

        memo[key] = res;
        return res;


    }
    vector<int> getCurrentStickerCount(vector<string>& stickers, int idx){
        string currentSticker = stickers[idx];
        if(stickerLetterCounts[idx][26] == 0){
            stickerLetterCounts[idx][26] = 1;
            for(char c:currentSticker){
                stickerLetterCounts[idx][c-'a']++;
            }
        }
        return stickerLetterCounts[idx];
    }
};
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