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leetcode-problemset/leetcode/originData/compare-strings-by-frequency-of-the-smallest-character.json
2022-05-02 23:44:12 +08:00

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{
"data": {
"question": {
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"title": "Compare Strings by Frequency of the Smallest Character",
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"content": "<p>Let the function <code>f(s)</code> be the <strong>frequency of the lexicographically smallest character</strong> in a non-empty string <code>s</code>. For example, if <code>s = &quot;dcce&quot;</code> then <code>f(s) = 2</code> because the lexicographically smallest character is <code>&#39;c&#39;</code>, which has a frequency of 2.</p>\n\n<p>You are given an array of strings <code>words</code> and another array of query strings <code>queries</code>. For each query <code>queries[i]</code>, count the <strong>number of words</strong> in <code>words</code> such that <code>f(queries[i])</code> &lt; <code>f(W)</code> for each <code>W</code> in <code>words</code>.</p>\n\n<p>Return <em>an integer array </em><code>answer</code><em>, where each </em><code>answer[i]</code><em> is the answer to the </em><code>i<sup>th</sup></code><em> query</em>.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> queries = [&quot;cbd&quot;], words = [&quot;zaaaz&quot;]\n<strong>Output:</strong> [1]\n<strong>Explanation:</strong> On the first query we have f(&quot;cbd&quot;) = 1, f(&quot;zaaaz&quot;) = 3 so f(&quot;cbd&quot;) &lt; f(&quot;zaaaz&quot;).\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> queries = [&quot;bbb&quot;,&quot;cc&quot;], words = [&quot;a&quot;,&quot;aa&quot;,&quot;aaa&quot;,&quot;aaaa&quot;]\n<strong>Output:</strong> [1,2]\n<strong>Explanation:</strong> On the first query only f(&quot;bbb&quot;) &lt; f(&quot;aaaa&quot;). On the second query both f(&quot;aaa&quot;) and f(&quot;aaaa&quot;) are both &gt; f(&quot;cc&quot;).\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 &lt;= queries.length &lt;= 2000</code></li>\n\t<li><code>1 &lt;= words.length &lt;= 2000</code></li>\n\t<li><code>1 &lt;= queries[i].length, words[i].length &lt;= 10</code></li>\n\t<li><code>queries[i][j]</code>, <code>words[i][j]</code> consist of lowercase English letters.</li>\n</ul>\n",
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"code": "class Solution {\npublic:\n vector<int> numSmallerByFrequency(vector<string>& queries, vector<string>& words) {\n \n }\n};",
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"code": "class Solution(object):\n def numSmallerByFrequency(self, queries, words):\n \"\"\"\n :type queries: List[str]\n :type words: List[str]\n :rtype: List[int]\n \"\"\"\n ",
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"code": "\n\n/**\n * Note: The returned array must be malloced, assume caller calls free().\n */\nint* numSmallerByFrequency(char ** queries, int queriesSize, char ** words, int wordsSize, int* returnSize){\n\n}",
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"code": "/**\n * @param {string[]} queries\n * @param {string[]} words\n * @return {number[]}\n */\nvar numSmallerByFrequency = function(queries, words) {\n \n};",
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"code": "(define/contract (num-smaller-by-frequency queries words)\n (-> (listof string?) (listof string?) (listof exact-integer?))\n\n )",
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"hints": [
"For each string from words calculate the leading count and store it in an array, then sort the integer array.",
"For each string from queries calculate the leading count \"p\" and in base of the sorted array calculated on the step 1 do a binary search to count the number of items greater than \"p\"."
],
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"sampleTestCase": "[\"cbd\"]\n[\"zaaaz\"]",
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