Class Solution
java.lang.Object
g2101_2200.s2135_count_words_obtained_after_adding_a_letter.Solution
2135 - Count Words Obtained After Adding a Letter.<p>Medium</p>
<p>You are given two <strong>0-indexed</strong> arrays of strings <code>startWords</code> and <code>targetWords</code>. Each string consists of <strong>lowercase English letters</strong> only.</p>
<p>For each string in <code>targetWords</code>, check if it is possible to choose a string from <code>startWords</code> and perform a <strong>conversion operation</strong> on it to be equal to that from <code>targetWords</code>.</p>
<p>The <strong>conversion operation</strong> is described in the following two steps:</p>
<ol>
<li><strong>Append</strong> any lowercase letter that is <strong>not present</strong> in the string to its end.
<ul>
<li>For example, if the string is <code>"abc"</code>, the letters <code>'d'</code>, <code>'e'</code>, or <code>'y'</code> can be added to it, but not <code>'a'</code>. If <code>'d'</code> is added, the resulting string will be <code>"abcd"</code>.</li>
</ul>
</li>
<li><strong>Rearrange</strong> the letters of the new string in <strong>any</strong> arbitrary order.
<ul>
<li>For example, <code>"abcd"</code> can be rearranged to <code>"acbd"</code>, <code>"bacd"</code>, <code>"cbda"</code>, and so on. Note that it can also be rearranged to <code>"abcd"</code> itself.</li>
</ul>
</li>
</ol>
<p>Return <em>the <strong>number of strings</strong> in</em> <code>targetWords</code> <em>that can be obtained by performing the operations on <strong>any</strong> string of</em> <code>startWords</code>.</p>
<p><strong>Note</strong> that you will only be verifying if the string in <code>targetWords</code> can be obtained from a string in <code>startWords</code> by performing the operations. The strings in <code>startWords</code> <strong>do not</strong> actually change during this process.</p>
<p><strong>Example 1:</strong></p>
<p><strong>Input:</strong> startWords = [“ant”,“act”,“tack”], targetWords = [“tack”,“act”,“acti”]</p>
<p><strong>Output:</strong> 2</p>
<p><strong>Explanation:</strong></p>
<ul>
<li>
<p>In order to form targetWords[0] = “tack”, we use startWords[1] = “act”, append ‘k’ to it, and rearrange “actk” to “tack”.</p>
</li>
<li>
<p>There is no string in startWords that can be used to obtain targetWords[1] = “act”.</p>
</li>
</ul>
<p>Note that “act” does exist in startWords, but we <strong>must</strong> append one letter to the string before rearranging it.</p>
<ul>
<li>In order to form targetWords[2] = “acti”, we use startWords[1] = “act”, append ‘i’ to it, and rearrange “acti” to “acti” itself.</li>
</ul>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> startWords = [“ab”,“a”], targetWords = [“abc”,“abcd”]</p>
<p><strong>Output:</strong> 1</p>
<p><strong>Explanation:</strong></p>
<ul>
<li>
<p>In order to form targetWords[0] = “abc”, we use startWords[0] = “ab”, add ‘c’ to it, and rearrange it to “abc”.</p>
</li>
<li>
<p>There is no string in startWords that can be used to obtain targetWords[1] = “abcd”.</p>
</li>
</ul>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>1 <= startWords.length, targetWords.length <= 5 * 10<sup>4</sup></code></li>
<li><code>1 <= startWords[i].length, targetWords[j].length <= 26</code></li>
<li>Each string of <code>startWords</code> and <code>targetWords</code> consists of lowercase English letters only.</li>
<li>No letter occurs more than once in any string of <code>startWords</code> or <code>targetWords</code>.</li>
</ul>
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public Solution()
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