number-of-valid-words-for-each-puzzle

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Algorithms

With respect to a given <code>puzzle</code> string, a <code>word</code> is valid if both the following conditions are satisfied:
<ul>
	<li><code>word</code> contains the first letter of <code>puzzle</code>.</li>
	<li>For each letter in <code>word</code>, that letter is in <code>puzzle</code>.
	<ul>
		<li>For example, if the puzzle is <code>&quot;abcdefg&quot;</code>, then valid words are <code>&quot;faced&quot;</code>, <code>&quot;cabbage&quot;</code>, and <code>&quot;baggage&quot;</code>, while</li>
		<li>invalid words are <code>&quot;beefed&quot;</code> (does not include <code>&#39;a&#39;</code>) and <code>&quot;based&quot;</code> (includes <code>&#39;s&#39;</code> which is not in the puzzle).</li>
	</ul>
	</li>
</ul>
Return an array <code>answer</code>, where <code>answer[i]</code> is the number of words in the given word list <code>words</code> that is valid with respect to the puzzle <code>puzzles[i]</code>.
&nbsp;
Example 1:

<pre>
Input: words = [&quot;aaaa&quot;,&quot;asas&quot;,&quot;able&quot;,&quot;ability&quot;,&quot;actt&quot;,&quot;actor&quot;,&quot;access&quot;], puzzles = [&quot;aboveyz&quot;,&quot;abrodyz&quot;,&quot;abslute&quot;,&quot;absoryz&quot;,&quot;actresz&quot;,&quot;gaswxyz&quot;]
Output: [1,1,3,2,4,0]
Explanation: 
1 valid word for &quot;aboveyz&quot; : &quot;aaaa&quot; 
1 valid word for &quot;abrodyz&quot; : &quot;aaaa&quot;
3 valid words for &quot;abslute&quot; : &quot;aaaa&quot;, &quot;asas&quot;, &quot;able&quot;
2 valid words for &quot;absoryz&quot; : &quot;aaaa&quot;, &quot;asas&quot;
4 valid words for &quot;actresz&quot; : &quot;aaaa&quot;, &quot;asas&quot;, &quot;actt&quot;, &quot;access&quot;
There are no valid words for &quot;gaswxyz&quot; cause none of the words in the list contains letter &#39;g&#39;.
</pre>

Example 2:

<pre>
Input: words = [&quot;apple&quot;,&quot;pleas&quot;,&quot;please&quot;], puzzles = [&quot;aelwxyz&quot;,&quot;aelpxyz&quot;,&quot;aelpsxy&quot;,&quot;saelpxy&quot;,&quot;xaelpsy&quot;]
Output: [0,1,3,2,0]
</pre>

&nbsp;
Constraints:

<ul>
	<li><code>1 &lt;= words.length &lt;= 105</code></li>
	<li><code>4 &lt;= words[i].length &lt;= 50</code></li>
	<li><code>1 &lt;= puzzles.length &lt;= 104</code></li>
	<li><code>puzzles[i].length == 7</code></li>
	<li><code>words[i]</code> and <code>puzzles[i]</code> consist of lowercase English letters.</li>
	<li>Each <code>puzzles[i] </code>does not contain repeated characters.</li>
</ul>

findNumOfValidWords

Number of Valid Words for Each Puzzle

Bit Manipulation

Trie

Array

Hash Table

String

外国友人仿照中国字谜设计了一个英文版猜字谜小游戏，请你来猜猜看吧。

字谜的迷面 <code>puzzle</code> 按字符串形式给出，如果一个单词 <code>word</code> 符合下面两个条件，那么它就可以算作谜底：

<ul>
	<li>单词 <code>word</code> 中包含谜面 <code>puzzle</code> 的第一个字母。</li>
	<li>单词 <code>word</code> 中的每一个字母都可以在谜面 <code>puzzle</code> 中找到。 
	例如，如果字谜的谜面是 "abcdefg"，那么可以作为谜底的单词有 "faced", "cabbage", 和 "baggage"；而 "beefed"（不含字母 "a"）以及 "based"（其中的 "s" 没有出现在谜面中）都不能作为谜底。</li>
</ul>

返回一个答案数组 <code>answer</code>，数组中的每个元素 <code>answer[i]</code> 是在给出的单词列表 <code>words</code> 中可以作为字谜迷面 <code>puzzles[i]</code> 所对应的谜底的单词数目。

 

示例：

<pre>
输入：
words = ["aaaa","asas","able","ability","actt","actor","access"], 
puzzles = ["aboveyz","abrodyz","abslute","absoryz","actresz","gaswxyz"]
输出：[1,1,3,2,4,0]
解释：
1 个单词可以作为 "aboveyz" 的谜底 : "aaaa" 
1 个单词可以作为 "abrodyz" 的谜底 : "aaaa"
3 个单词可以作为 "abslute" 的谜底 : "aaaa", "asas", "able"
2 个单词可以作为 "absoryz" 的谜底 : "aaaa", "asas"
4 个单词可以作为 "actresz" 的谜底 : "aaaa", "asas", "actt", "access"
没有单词可以作为 "gaswxyz" 的谜底，因为列表中的单词都不含字母 'g'。
</pre>

 

提示：

<ul>
	<li><code>1 <= words.length <= 10^5</code></li>
	<li><code>4 <= words[i].length <= 50</code></li>
	<li><code>1 <= puzzles.length <= 10^4</code></li>
	<li><code>puzzles[i].length == 7</code></li>
	<li><code>words[i][j]</code>, <code>puzzles[i][j]</code> 都是小写英文字母。</li>
	<li>每个 <code>puzzles[i]</code> 所包含的字符都不重复。</li>
</ul>