<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:
    int minStickers(vector<string>& s, string t) {
        int size = t.size();
        vector<int> v(1 << size, 0);
        queue<int> q;
        int round = 0;
        q.push(0);
        v[0] = 1;
        while (!q.empty()) {
            int qsize = q.size();
            ++round;
            for (int i = 0; i < qsize; ++i) {
                int front = q.front();
                q.pop();
                for (string& tmp : s) {
                    vector<int> num(26, 0);
                    for (char c : tmp) {
                        num[c - 'a']++;
                    }
                    int status = front;
                    for (int j = 0; j < size; ++j) {
                        if ((status & (1 << j)) == 0 && num[t[j] - 'a'] > 0) {
                            status |= (1 << j);
                            num[t[j] - 'a']--;
                        }
                    }
                    if (status == (1 << size) - 1) {
                        return round;
                    }
                    if (v[status] == 0) {
                        v[status] = 1;
                        q.push(status);
                    }
                }
            }
        }   
        return -1;
    }
};

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