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"question": {
"questionId": "2186",
"questionFrontendId": "2062",
"categoryTitle": "Algorithms",
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"title": "Count Vowel Substrings of a String",
"titleSlug": "count-vowel-substrings-of-a-string",
"content": "<p>A <strong>substring</strong> is a contiguous (non-empty) sequence of characters within a string.</p>\n\n<p>A <strong>vowel substring</strong> is a substring that <strong>only</strong> consists of vowels (<code>&#39;a&#39;</code>, <code>&#39;e&#39;</code>, <code>&#39;i&#39;</code>, <code>&#39;o&#39;</code>, and <code>&#39;u&#39;</code>) and has <strong>all five</strong> vowels present in it.</p>\n\n<p>Given a string <code>word</code>, return <em>the number of <strong>vowel substrings</strong> in</em> <code>word</code>.</p>\n\n<p>&nbsp;</p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;aeiouu&quot;\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The vowel substrings of word are as follows (underlined):\n- &quot;<strong><u>aeiou</u></strong>u&quot;\n- &quot;<strong><u>aeiouu</u></strong>&quot;\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;unicornarihan&quot;\n<strong>Output:</strong> 0\n<strong>Explanation:</strong> Not all 5 vowels are present, so there are no vowel substrings.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> word = &quot;cuaieuouac&quot;\n<strong>Output:</strong> 7\n<strong>Explanation:</strong> The vowel substrings of word are as follows (underlined):\n- &quot;c<strong><u>uaieuo</u></strong>uac&quot;\n- &quot;c<strong><u>uaieuou</u></strong>ac&quot;\n- &quot;c<strong><u>uaieuoua</u></strong>c&quot;\n- &quot;cu<strong><u>aieuo</u></strong>uac&quot;\n- &quot;cu<strong><u>aieuou</u></strong>ac&quot;\n- &quot;cu<strong><u>aieuoua</u></strong>c&quot;\n- &quot;cua<strong><u>ieuoua</u></strong>c&quot;\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 100</code></li>\n\t<li><code>word</code> consists of lowercase English letters only.</li>\n</ul>\n",
"translatedTitle": "统计字符串中的元音子字符串",
"translatedContent": "<p><strong>子字符串</strong> 是字符串中的一个连续(非空)的字符序列。</p>\n\n<p><strong>元音子字符串</strong> 是 <strong>仅</strong> 由元音(<code>'a'</code>、<code>'e'</code>、<code>'i'</code>、<code>'o'</code> 和 <code>'u'</code>)组成的一个子字符串,且必须包含 <strong>全部五种</strong> 元音。</p>\n\n<p>给你一个字符串 <code>word</code> ,统计并返回 <code>word</code> 中 <strong>元音子字符串的数目</strong> 。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1</strong></p>\n\n<pre>\n<strong>输入:</strong>word = \"aeiouu\"\n<strong>输出:</strong>2\n<strong>解释:</strong>下面列出 word 中的元音子字符串(斜体加粗部分):\n- \"<em><strong>aeiou</strong></em>u\"\n- \"<strong><em>aeiouu</em></strong>\"\n</pre>\n\n<p><strong>示例 2</strong></p>\n\n<pre>\n<strong>输入:</strong>word = \"unicornarihan\"\n<strong>输出:</strong>0\n<strong>解释:</strong>word 中不含 5 种元音,所以也不会存在元音子字符串。\n</pre>\n\n<p><strong>示例 3</strong></p>\n\n<pre>\n<strong>输入:</strong>word = \"cuaieuouac\"\n<strong>输出:</strong>7\n<strong>解释:</strong>下面列出 word 中的元音子字符串(斜体加粗部分):\n- \"c<em><strong>uaieuo</strong></em>uac\"\n- \"c<em><strong>uaieuou</strong></em>ac\"\n- \"c<em><strong>uaieuoua</strong></em>c\"\n- \"cu<em><strong>aieuo</strong></em>uac\"\n- \"cu<em><strong>aieuou</strong></em>ac\"\n- \"cu<em><strong>aieuoua</strong></em>c\"\n- \"cua<em><strong>ieuoua</strong></em>c\"</pre>\n\n<p><strong>示例 4</strong></p>\n\n<pre>\n<strong>输入:</strong>word = \"bbaeixoubb\"\n<strong>输出:</strong>0\n<strong>解释:</strong>所有包含全部五种元音的子字符串都含有辅音,所以不存在元音子字符串。\n</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= word.length &lt;= 100</code></li>\n\t<li><code>word</code> 仅由小写英文字母组成</li>\n</ul>\n",
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"While generating substrings starting at any index, do you need to continue generating larger substrings if you encounter a consonant?",
"Can you store the count of characters to avoid generating substrings altogether?"
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