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leetcode-problemset/leetcode/originData/count-words-obtained-after-adding-a-letter.json

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{
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"title": "Count Words Obtained After Adding a Letter",
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"content": "<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>\n\n<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>\n\n<p>The <strong>conversion operation</strong> is described in the following two steps:</p>\n\n<ol>\n\t<li><strong>Append</strong> any lowercase letter that is <strong>not present</strong> in the string to its end.\n\n\t<ul>\n\t\t<li>For example, if the string is <code>&quot;abc&quot;</code>, the letters <code>&#39;d&#39;</code>, <code>&#39;e&#39;</code>, or <code>&#39;y&#39;</code> can be added to it, but not <code>&#39;a&#39;</code>. If <code>&#39;d&#39;</code> is added, the resulting string will be <code>&quot;abcd&quot;</code>.</li>\n\t</ul>\n\t</li>\n\t<li><strong>Rearrange</strong> the letters of the new string in <strong>any</strong> arbitrary order.\n\t<ul>\n\t\t<li>For example, <code>&quot;abcd&quot;</code> can be rearranged to <code>&quot;acbd&quot;</code>, <code>&quot;bacd&quot;</code>, <code>&quot;cbda&quot;</code>, and so on. Note that it can also be rearranged to <code>&quot;abcd&quot;</code> itself.</li>\n\t</ul>\n\t</li>\n</ol>\n\n<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>\n\n<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>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> startWords = [&quot;ant&quot;,&quot;act&quot;,&quot;tack&quot;], targetWords = [&quot;tack&quot;,&quot;act&quot;,&quot;acti&quot;]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\n- In order to form targetWords[0] = &quot;tack&quot;, we use startWords[1] = &quot;act&quot;, append &#39;k&#39; to it, and rearrange &quot;actk&quot; to &quot;tack&quot;.\n- There is no string in startWords that can be used to obtain targetWords[1] = &quot;act&quot;.\n Note that &quot;act&quot; does exist in startWords, but we <strong>must</strong> append one letter to the string before rearranging it.\n- In order to form targetWords[2] = &quot;acti&quot;, we use startWords[1] = &quot;act&quot;, append &#39;i&#39; to it, and rearrange &quot;acti&quot; to &quot;acti&quot; itself.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> startWords = [&quot;ab&quot;,&quot;a&quot;], targetWords = [&quot;abc&quot;,&quot;abcd&quot;]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>\n- In order to form targetWords[0] = &quot;abc&quot;, we use startWords[0] = &quot;ab&quot;, add &#39;c&#39; to it, and rearrange it to &quot;abc&quot;.\n- There is no string in startWords that can be used to obtain targetWords[1] = &quot;abcd&quot;.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= startWords.length, targetWords.length &lt;= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 &lt;= startWords[i].length, targetWords[j].length &lt;= 26</code></li>\n\t<li>Each string of <code>startWords</code> and <code>targetWords</code> consists of lowercase English letters only.</li>\n\t<li>No letter occurs more than once in any string of <code>startWords</code> or <code>targetWords</code>.</li>\n</ul>\n",
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"code": "class Solution {\npublic:\n int wordCount(vector<string>& startWords, vector<string>& targetWords) {\n \n }\n};",
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"Which data structure can be used to efficiently check if a string exists in startWords?",
"After appending a letter, all letters of a string can be rearranged in any possible way. How can we use this to reduce our search space while checking if a string in targetWords can be obtained from a string in startWords?"
],
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