Stickers to Spell Word
<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>
Bit Manipulation; Memoization; Array; Hash Table; String; Dynamic Programming; Backtracking; Bitmask
class Solution {
    public int minStickers(String[] stickers, String target) {
        int n = target.length();
        Deque<Integer> q = new ArrayDeque<>();
        q.offer(0);
        boolean[] vis = new boolean[1 << n];
        vis[0] = true;
        for (int ans = 0; !q.isEmpty(); ++ans) {
            for (int m = q.size(); m > 0; --m) {
                int cur = q.poll();
                if (cur == (1 << n) - 1) {
                    return ans;
                }
                for (String s : stickers) {
                    int[] cnt = new int[26];
                    int nxt = cur;
                    for (char c : s.toCharArray()) {
                        ++cnt[c - 'a'];
                    }
                    for (int i = 0; i < n; ++i) {
                        int j = target.charAt(i) - 'a';
                        if ((cur >> i & 1) == 0 && cnt[j] > 0) {
                            --cnt[j];
                            nxt |= 1 << i;
                        }
                    }
                    if (!vis[nxt]) {
                        vis[nxt] = true;
                        q.offer(nxt);
                    }
                }
            }
        }
        return -1;
    }
}