<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 find(int i, int targetMask, int m, int n, vector<string>& stickers,
             string& target, vector<vector<int>>& dp) {

        if (targetMask == ((1 << m) - 1))
            return 0;

        if (i == n)
            return INT_MAX;

        if (dp[i][targetMask] != -1)
            return dp[i][targetMask];

        int notTake, take = INT_MAX;

        vector<int> freq(26, 0);
        for (auto c : stickers[i])
            freq[c - 'a']++;

        int newTargetMask = 0;
        for (int j = 0; j < m; j++) {
            int maskPos = m - j - 1;
            if ((targetMask & (1 << maskPos)) == 0) {
                if (freq[target[j] - 'a'] > 0) {
                    freq[target[j] - 'a']--;
                    newTargetMask |= 1 << maskPos;
                }
            }
        }

        if (newTargetMask != 0)
            take =
                find(i, targetMask | newTargetMask, m, n, stickers, target, dp);
        if (take != INT_MAX)
            take++;

        notTake = find(i + 1, targetMask, m, n, stickers, target, dp);

        return dp[i][targetMask] = min(notTake, take);
    }

    int minStickers(vector<string>& stickers, string target) {
        int n = stickers.size(), m = target.length();
        int targetMask = 0;

        vector<vector<int>> dp(n, vector<int>(1 << m, -1));
        int ans = find(0, targetMask, m, n, stickers, target, dp);
        return ans == INT_MAX ? -1 : ans;
    }
};
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