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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>"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>\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>"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>\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> </p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> startWords = ["ant","act","tack"], targetWords = ["tack","act","acti"]\n<strong>Output:</strong> 2\n<strong>Explanation:</strong>\n- In order to form targetWords[0] = "tack", we use startWords[1] = "act", append 'k' to it, and rearrange "actk" to "tack".\n- There is no string in startWords that can be used to obtain targetWords[1] = "act".\n Note that "act" 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] = "acti", we use startWords[1] = "act", append 'i' to it, and rearrange "acti" to "acti" itself.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> startWords = ["ab","a"], targetWords = ["abc","abcd"]\n<strong>Output:</strong> 1\n<strong>Explanation:</strong>\n- In order to form targetWords[0] = "abc", we use startWords[0] = "ab", add 'c' to it, and rearrange it to "abc".\n- There is no string in startWords that can be used to obtain targetWords[1] = "abcd".\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= startWords.length, targetWords.length <= 5 * 10<sup>4</sup></code></li>\n\t<li><code>1 <= startWords[i].length, targetWords[j].length <= 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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"code": "class Solution(object):\n def wordCount(self, startWords, targetWords):\n \"\"\"\n :type startWords: List[str]\n :type targetWords: List[str]\n :rtype: int\n \"\"\"\n ",
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"code": "/**\n * @param {string[]} startWords\n * @param {string[]} targetWords\n * @return {number}\n */\nvar wordCount = function(startWords, targetWords) {\n \n};",
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"code": "# @param {String[]} start_words\n# @param {String[]} target_words\n# @return {Integer}\ndef word_count(start_words, target_words)\n \nend",
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"lang": "PHP",
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"code": "(define/contract (word-count startWords targetWords)\n (-> (listof string?) (listof string?) exact-integer?)\n\n )",
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"lang": "Erlang",
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"code": "-spec word_count(StartWords :: [unicode:unicode_binary()], TargetWords :: [unicode:unicode_binary()]) -> integer().\nword_count(StartWords, TargetWords) ->\n .",
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"code": "defmodule Solution do\n @spec word_count(start_words :: [String.t], target_words :: [String.t]) :: integer\n def word_count(start_words, target_words) do\n\n end\nend",
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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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"sampleTestCase": "[\"ant\",\"act\",\"tack\"]\n[\"tack\",\"act\",\"acti\"]",
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Adding an item to a hash.</b>\\r\\n<pre>\\r\\nstruct hash_entry {\\r\\n int id; /* we'll use this field as the key */\\r\\n char name[10];\\r\\n UT_hash_handle hh; /* makes this structure hashable */\\r\\n};\\r\\n\\r\\nstruct hash_entry *users = NULL;\\r\\n\\r\\nvoid add_user(struct hash_entry *s) {\\r\\n HASH_ADD_INT(users, id, s);\\r\\n}\\r\\n</pre>\\r\\n</p>\\r\\n\\r\\n<p><b>2. Looking up an item in a hash:</b>\\r\\n<pre>\\r\\nstruct hash_entry *find_user(int user_id) {\\r\\n struct hash_entry *s;\\r\\n HASH_FIND_INT(users, &user_id, s);\\r\\n return s;\\r\\n}\\r\\n</pre>\\r\\n</p>\\r\\n\\r\\n<p><b>3. 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