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leetcode-problemset/leetcode/originData/check-if-a-string-is-an-acronym-of-words.json

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{
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"title": "Check if a String Is an Acronym of Words",
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"content": "<p>Given an array of strings <code>words</code> and a string <code>s</code>, determine if <code>s</code> is an <strong>acronym</strong> of words.</p>\n\n<p>The string <code>s</code> is considered an acronym of <code>words</code> if it can be formed by concatenating the <strong>first</strong> character of each string in <code>words</code> <strong>in order</strong>. For example, <code>&quot;ab&quot;</code> can be formed from <code>[&quot;apple&quot;, &quot;banana&quot;]</code>, but it can&#39;t be formed from <code>[&quot;bear&quot;, &quot;aardvark&quot;]</code>.</p>\n\n<p>Return <code>true</code><em> if </em><code>s</code><em> is an acronym of </em><code>words</code><em>, and </em><code>false</code><em> otherwise. </em></p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> words = [&quot;alice&quot;,&quot;bob&quot;,&quot;charlie&quot;], s = &quot;abc&quot;\n<strong>Output:</strong> true\n<strong>Explanation:</strong> The first character in the words &quot;alice&quot;, &quot;bob&quot;, and &quot;charlie&quot; are &#39;a&#39;, &#39;b&#39;, and &#39;c&#39;, respectively. Hence, s = &quot;abc&quot; is the acronym. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> words = [&quot;an&quot;,&quot;apple&quot;], s = &quot;a&quot;\n<strong>Output:</strong> false\n<strong>Explanation:</strong> The first character in the words &quot;an&quot; and &quot;apple&quot; are &#39;a&#39; and &#39;a&#39;, respectively. \nThe acronym formed by concatenating these characters is &quot;aa&quot;. \nHence, s = &quot;a&quot; is not the acronym.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> words = [&quot;never&quot;,&quot;gonna&quot;,&quot;give&quot;,&quot;up&quot;,&quot;on&quot;,&quot;you&quot;], s = &quot;ngguoy&quot;\n<strong>Output:</strong> true\n<strong>Explanation: </strong>By concatenating the first character of the words in the array, we get the string &quot;ngguoy&quot;. \nHence, s = &quot;ngguoy&quot; is the acronym.\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= words.length &lt;= 100</code></li>\n\t<li><code>1 &lt;= words[i].length &lt;= 10</code></li>\n\t<li><code>1 &lt;= s.length &lt;= 100</code></li>\n\t<li><code>words[i]</code> and <code>s</code> consist of lowercase English letters.</li>\n</ul>\n",
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"code": "class Solution(object):\n def isAcronym(self, words, s):\n \"\"\"\n :type words: List[str]\n :type s: str\n :rtype: bool\n \"\"\"\n ",
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"code": "defmodule Solution do\n @spec is_acronym(words :: [String.t], s :: String.t) :: boolean\n def is_acronym(words, s) do\n\n end\nend",
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"hints": [
"<div class=\"_1l1MA\">Concatenate the first characters of the strings in <code>words</code>, and compare the resulting concatenation to <code>s</code>.</div>"
],
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"sampleTestCase": "[\"alice\",\"bob\",\"charlie\"]\n\"abc\"",
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