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README.md
leetcode-cn
origin-data.json
originData
[no content]count-strictly-increasing-subarrays.json[no content]employees-with-deductions.json[no content]median-of-a-row-wise-sorted-matrix.jsonfind-subarrays-with-equal-sum.jsonfind-the-index-of-the-first-occurrence-in-a-string.jsonmaximum-number-of-robots-within-budget.jsonmaximum-rows-covered-by-columns.jsonstrictly-palindromic-number.json
problem (Chinese)
严格回文的数字 [strictly-palindromic-number].html和相等的子数组 [find-subarrays-with-equal-sum].html实现 strStr() [find-the-index-of-the-first-occurrence-in-a-string].html被列覆盖的最多行数 [maximum-rows-covered-by-columns].html预算内的最多机器人数目 [maximum-number-of-robots-within-budget].html
problem (English)
严格回文的数字(English) [strictly-palindromic-number].html和相等的子数组(English) [find-subarrays-with-equal-sum].html实现 strStr()(English) [find-the-index-of-the-first-occurrence-in-a-string].html被列覆盖的最多行数(English) [maximum-rows-covered-by-columns].html预算内的最多机器人数目(English) [maximum-number-of-robots-within-budget].html
leetcode
@ -1,6 +1,6 @@
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# 力扣题库（完整版）
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> 最后更新日期： **2022.08.29**
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> 最后更新日期： **2022.09.04**
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>
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> 使用脚本前请务必仔细完整阅读本 `README.md` 文件
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||||
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{
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"data": {
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||||
"question": {
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||||
"questionId": "2534",
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||||
"questionFrontendId": "2394",
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||||
"categoryTitle": "Database",
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||||
"boundTopicId": 1795924,
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"title": "Employees With Deductions",
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||||
"titleSlug": "employees-with-deductions",
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"content": null,
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"translatedTitle": null,
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||||
"translatedContent": null,
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||||
"isPaidOnly": true,
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||||
"difficulty": "Medium",
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||||
"likes": 0,
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"dislikes": 0,
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"isLiked": null,
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"similarQuestions": "[]",
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"contributors": [],
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"langToValidPlayground": null,
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||||
"topicTags": [],
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"companyTagStats": null,
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"codeSnippets": null,
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"stats": "{\"totalAccepted\": \"12\", \"totalSubmission\": \"15\", \"totalAcceptedRaw\": 12, \"totalSubmissionRaw\": 15, \"acRate\": \"80.0%\"}",
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"hints": [],
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"solution": null,
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"status": null,
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"sampleTestCase": "{\"headers\": {\"Employees\": [\"employee_id\", \"needed_hours\"], \"Logs\": [\"employee_id\", \"in_time\", \"out_time\"]}, \"rows\": {\"Employees\": [[1, 20], [2, 12], [3, 2]], \"Logs\": [[1, \"2022-10-01 09:00:00\", \"2022-10-01 17:00:00\"],[1, \"2022-10-06 09:05:04\", \"2022-10-06 17:09:03\"], [1, \"2022-10-12 23:00:00\", \"2022-10-13 03:00:01\"], [2, \"2022-10-29 12:00:00\", \"2022-10-29 23:58:58\"]]}}",
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"metaData": "{\n \"mysql\": [\n \"Create table If Not Exists Employees (employee_id int, needed_hours int)\",\n \"Create table If Not Exists Logs (employee_id int, in_time datetime, out_time datetime)\"\n ],\n \"mssql\": [\n \"Create table Employees (employee_id int, needed_hours int)\",\n \"Create table Logs (employee_id int, in_time datetime, out_time datetime)\"\n ],\n \"oraclesql\": [\n \"Create table Employees (employee_id int, needed_hours int)\",\n \"Create table Logs (employee_id int, in_time date, out_time date)\",\n \"ALTER SESSION SET nls_date_format='YYYY-MM-DD HH24:MI:SS'\"\n ],\n \"database\": true\n}",
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"judgerAvailable": true,
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"judgeType": "large",
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"mysqlSchemas": [
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"Create table If Not Exists Employees (employee_id int, needed_hours int)",
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"Create table If Not Exists Logs (employee_id int, in_time datetime, out_time datetime)",
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"Truncate table Employees",
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"insert into Employees (employee_id, needed_hours) values ('1', '20')",
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"insert into Employees (employee_id, needed_hours) values ('2', '12')",
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"insert into Employees (employee_id, needed_hours) values ('3', '2')",
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"Truncate table Logs",
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"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-01 09:00:00', '2022-10-01 17:00:00')",
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"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-06 09:05:04', '2022-10-06 17:09:03')",
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"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-12 23:00:00', '2022-10-13 03:00:01')",
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"insert into Logs (employee_id, in_time, out_time) values ('2', '2022-10-29 12:00:00', '2022-10-29 23:58:58')"
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],
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"enableRunCode": true,
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"envInfo": "{\"mysql\":[\"MySQL\",\"<p>\\u7248\\u672c\\uff1a<code>MySQL 8.0<\\/code><\\/p>\"],\"mssql\":[\"MS SQL Server\",\"<p>mssql server 2019.<\\/p>\"],\"oraclesql\":[\"Oracle\",\"<p>Oracle Sql 11.2.<\\/p>\"]}",
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"book": null,
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"isSubscribed": false,
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"isDailyQuestion": false,
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"dailyRecordStatus": null,
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"editorType": "CKEDITOR",
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"ugcQuestionId": null,
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"style": "LEETCODE",
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"exampleTestcases": "{\"headers\": {\"Employees\": [\"employee_id\", \"needed_hours\"], \"Logs\": [\"employee_id\", \"in_time\", \"out_time\"]}, \"rows\": {\"Employees\": [[1, 20], [2, 12], [3, 2]], \"Logs\": [[1, \"2022-10-01 09:00:00\", \"2022-10-01 17:00:00\"],[1, \"2022-10-06 09:05:04\", \"2022-10-06 17:09:03\"], [1, \"2022-10-12 23:00:00\", \"2022-10-13 03:00:01\"], [2, \"2022-10-29 12:00:00\", \"2022-10-29 23:58:58\"]]}}",
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"__typename": "QuestionNode"
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}
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}
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}
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164
leetcode-cn/originData/find-subarrays-with-equal-sum.json
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leetcode-cn/originData/find-subarrays-with-equal-sum.json
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leetcode-cn/originData/maximum-rows-covered-by-columns.json
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leetcode-cn/originData/maximum-rows-covered-by-columns.json
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leetcode-cn/originData/strictly-palindromic-number.json
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leetcode-cn/originData/strictly-palindromic-number.json
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<p>如果一个整数 <code>n</code> 在 <code>b</code> 进制下（<code>b</code> 为 <code>2</code> 到 <code>n - 2</code> 之间的所有整数）对应的字符串 <strong>全部</strong> 都是 <strong>回文的</strong> ，那么我们称这个数 <code>n</code> 是 <strong>严格回文</strong> 的。</p>
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<p>给你一个整数 <code>n</code> ，如果 <code>n</code> 是 <strong>严格回文</strong> 的，请返回 <code>true</code> ，否则返回<em> </em><code>false</code> 。</p>
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<p>如果一个字符串从前往后读和从后往前读完全相同，那么这个字符串是 <strong>回文的</strong> 。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>n = 9
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<b>输出：</b>false
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<b>解释：</b>在 2 进制下：9 = 1001 ，是回文的。
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在 3 进制下：9 = 100 ，不是回文的。
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所以，9 不是严格回文数字，我们返回 false 。
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注意在 4, 5, 6 和 7 进制下，n = 9 都不是回文的。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>n = 4
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<b>输出：</b>false
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<b>解释：</b>我们只考虑 2 进制：4 = 100 ，不是回文的。
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所以我们返回 false 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>4 <= n <= 10<sup>5</sup></code></li>
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</ul>
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<p>给你一个下标从 <strong>0</strong> 开始的整数数组 <code>nums</code> ，判断是否存在 <strong>两个</strong> 长度为 <code>2</code> 的子数组且它们的 <strong>和</strong> 相等。注意，这两个子数组起始位置的下标必须 <strong>不相同</strong> 。</p>
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<p>如果这样的子数组存在，请返回 <code>true</code>，否则返回 <code>false</code><em> </em>。</p>
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<p><strong>子数组</strong> 是一个数组中一段连续非空的元素组成的序列。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre><b>输入：</b>nums = [4,2,4]
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<b>输出：</b>true
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<b>解释：</b>元素为 [4,2] 和 [2,4] 的子数组有相同的和 6 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre><b>输入：</b>nums = [1,2,3,4,5]
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<b>输出：</b>false
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<b>解释：</b>没有长度为 2 的两个子数组和相等。
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</pre>
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<p><strong>示例 3：</strong></p>
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<pre><b>输入：</b>nums = [0,0,0]
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<b>输出：</b>true
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<b>解释：</b>子数组 [nums[0],nums[1]] 和 [nums[1],nums[2]] 的和相等，都为 0 。
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注意即使子数组的元素相同，这两个子数组也视为不相同的子数组，因为它们在原数组中的起始位置不同。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>2 <= nums.length <= 1000</code></li>
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<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
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</ul>
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34
leetcode-cn/problem (Chinese)/实现 strStr() [find-the-index-of-the-first-occurrence-in-a-string].html
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leetcode-cn/problem (Chinese)/实现 strStr() [find-the-index-of-the-first-occurrence-in-a-string].html
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<p>实现 <a href="https://baike.baidu.com/item/strstr/811469" target="_blank">strStr()</a> 函数。</p>
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<p>给你两个字符串 <code>haystack</code> 和 <code>needle</code> ，请你在 <code>haystack</code> 字符串中找出 <code>needle</code> 字符串出现的第一个位置（下标从 0 开始）。如果不存在，则返回 <code>-1</code><strong> </strong>。</p>
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<p><strong>说明：</strong></p>
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<p>当 <code>needle</code> 是空字符串时，我们应当返回什么值呢？这是一个在面试中很好的问题。</p>
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<p>对于本题而言，当 <code>needle</code> 是空字符串时我们应当返回 0 。这与 C 语言的 <a href="https://baike.baidu.com/item/strstr/811469" target="_blank">strstr()</a> 以及 Java 的 <a href="https://docs.oracle.com/javase/7/docs/api/java/lang/String.html#indexOf(java.lang.String)" target="_blank">indexOf()</a> 定义相符。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<strong>输入：</strong>haystack = "hello", needle = "ll"
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<strong>输出：</strong>2
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<strong>输入：</strong>haystack = "aaaaa", needle = "bba"
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<strong>输出：</strong>-1
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>1 <= haystack.length, needle.length <= 10<sup>4</sup></code></li>
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<li><code>haystack</code> 和 <code>needle</code> 仅由小写英文字符组成</li>
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</ul>
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<p>给你一个下标从 <strong>0</strong> 开始的 <code>m x n</code> 二进制矩阵 <code>mat</code> 和一个整数 <code>cols</code> ，表示你需要选出的列数。</p>
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<p>如果一行中，所有的 <code>1</code> 都被你选中的列所覆盖，那么我们称这一行 <strong>被覆盖</strong> 了。</p>
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<p>请你返回在选择 <code>cols</code> 列的情况下，<strong>被覆盖</strong> 的行数 <strong>最大</strong> 为多少。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered.png" style="width: 250px; height: 417px;"></strong></p>
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<pre><b>输入：</b>mat = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], cols = 2
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<b>输出：</b>3
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<strong>解释：</strong>
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如上图所示，覆盖 3 行的一种可行办法是选择第 0 和第 2 列。
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可以看出，不存在大于 3 行被覆盖的方案，所以我们返回 3 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered2.png" style="width: 83px; height: 247px;"></strong></p>
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<pre><b>输入：</b>mat = [[1],[0]], cols = 1
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<b>输出：</b>2
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<strong>解释：</strong>
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选择唯一的一列，两行都被覆盖了，原因是整个矩阵都被覆盖了。
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所以我们返回 2 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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<ul>
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<li><code>m == mat.length</code></li>
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<li><code>n == mat[i].length</code></li>
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<li><code>1 <= m, n <= 12</code></li>
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<li><code>mat[i][j]</code> 要么是 <code>0</code> 要么是 <code>1</code> 。</li>
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<li><code>1 <= cols <= n</code></li>
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</ul>
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37
leetcode-cn/problem (Chinese)/预算内的最多机器人数目 [maximum-number-of-robots-within-budget].html
Normal file
37
leetcode-cn/problem (Chinese)/预算内的最多机器人数目 [maximum-number-of-robots-within-budget].html
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<p>你有 <code>n</code> 个机器人，给你两个下标从 <strong>0</strong> 开始的整数数组 <code>chargeTimes</code> 和 <code>runningCosts</code> ，两者长度都为 <code>n</code> 。第 <code>i</code> 个机器人充电时间为 <code>chargeTimes[i]</code> 单位时间，花费 <code>runningCosts[i]</code> 单位时间运行。再给你一个整数 <code>budget</code> 。</p>
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<p>运行 <code>k</code> 个机器人 <strong>总开销</strong> 是 <code>max(chargeTimes) + k * sum(runningCosts)</code> ，其中 <code>max(chargeTimes)</code> 是这 <code>k</code> 个机器人中最大充电时间，<code>sum(runningCosts)</code> 是这 <code>k</code> 个机器人的运行时间之和。</p>
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<p>请你返回在 <strong>不超过</strong> <code>budget</code> 的前提下，你 <strong>最多</strong> 可以 <strong>连续</strong> 运行的机器人数目为多少。</p>
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<p> </p>
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<p><strong>示例 1：</strong></p>
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<pre>
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<b>输入：</b>chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25
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<b>输出：</b>3
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<b>解释：</b>
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可以在 budget 以内运行所有单个机器人或者连续运行 2 个机器人。
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选择前 3 个机器人，可以得到答案最大值 3 。总开销是 max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 ，小于 25 。
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可以看出无法在 budget 以内连续运行超过 3 个机器人，所以我们返回 3 。
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</pre>
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<p><strong>示例 2：</strong></p>
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<pre>
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<b>输入：</b>chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19
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<b>输出：</b>0
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<b>解释：</b>即使运行任何一个单个机器人，还是会超出 budget，所以我们返回 0 。
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</pre>
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<p> </p>
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<p><strong>提示：</strong></p>
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||||
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||||
<ul>
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||||
<li><code>chargeTimes.length == runningCosts.length == n</code></li>
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<li><code>1 <= n <= 5 * 10<sup>4</sup></code></li>
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<li><code>1 <= chargeTimes[i], runningCosts[i] <= 10<sup>5</sup></code></li>
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<li><code>1 <= budget <= 10<sup>15</sup></code></li>
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</ul>
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<p>An integer <code>n</code> is <strong>strictly palindromic</strong> if, for <strong>every</strong> base <code>b</code> between <code>2</code> and <code>n - 2</code> (<strong>inclusive</strong>), the string representation of the integer <code>n</code> in base <code>b</code> is <strong>palindromic</strong>.</p>
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<p>Given an integer <code>n</code>, return <code>true</code> <em>if </em><code>n</code><em> is <strong>strictly palindromic</strong> and </em><code>false</code><em> otherwise</em>.</p>
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<p>A string is <strong>palindromic</strong> if it reads the same forward and backward.</p>
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<p> </p>
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||||
<p><strong>Example 1:</strong></p>
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||||
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<pre>
|
||||
<strong>Input:</strong> n = 9
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> In base 2: 9 = 1001 (base 2), which is palindromic.
|
||||
In base 3: 9 = 100 (base 3), which is not palindromic.
|
||||
Therefore, 9 is not strictly palindromic so we return false.
|
||||
Note that in bases 4, 5, 6, and 7, n = 9 is also not palindromic.
|
||||
</pre>
|
||||
|
||||
<p><strong>Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> n = 4
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> We only consider base 2: 4 = 100 (base 2), which is not palindromic.
|
||||
Therefore, we return false.
|
||||
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>4 <= n <= 10<sup>5</sup></code></li>
|
||||
</ul>
|
||||
@ -0,0 +1,39 @@
|
||||
<p>Given a <strong>0-indexed</strong> integer array <code>nums</code>, determine whether there exist <strong>two</strong> subarrays of length <code>2</code> with <strong>equal</strong> sum. Note that the two subarrays must begin at <strong>different</strong> indices.</p>
|
||||
|
||||
<p>Return <code>true</code><em> if these subarrays exist, and </em><code>false</code><em> otherwise.</em></p>
|
||||
|
||||
<p>A <b>subarray</b> is a contiguous non-empty sequence of elements within an array.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [4,2,4]
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> The subarrays with elements [4,2] and [2,4] have the same sum of 6.
|
||||
</pre>
|
||||
|
||||
<p><strong>Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [1,2,3,4,5]
|
||||
<strong>Output:</strong> false
|
||||
<strong>Explanation:</strong> No two subarrays of size 2 have the same sum.
|
||||
</pre>
|
||||
|
||||
<p><strong>Example 3:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> nums = [0,0,0]
|
||||
<strong>Output:</strong> true
|
||||
<strong>Explanation:</strong> The subarrays [nums[0],nums[1]] and [nums[1],nums[2]] have the same sum of 0.
|
||||
Note that even though the subarrays have the same content, the two subarrays are considered different because they are in different positions in the original array.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>2 <= nums.length <= 1000</code></li>
|
||||
<li><code>-10<sup>9</sup> <= nums[i] <= 10<sup>9</sup></code></li>
|
||||
</ul>
|
||||
27
leetcode-cn/problem (English)/实现 strStr()(English) [find-the-index-of-the-first-occurrence-in-a-string].html
Normal file
27
leetcode-cn/problem (English)/实现 strStr()(English) [find-the-index-of-the-first-occurrence-in-a-string].html
Normal file
@ -0,0 +1,27 @@
|
||||
<p>Given two strings <code>needle</code> and <code>haystack</code>, return the index of the first occurrence of <code>needle</code> in <code>haystack</code>, or <code>-1</code> if <code>needle</code> is not part of <code>haystack</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> haystack = "sadbutsad", needle = "sad"
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> "sad" occurs at index 0 and 6.
|
||||
The first occurrence is at index 0, so we return 0.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> haystack = "leetcode", needle = "leeto"
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> "leeto" did not occur in "leetcode", so we return -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= haystack.length, needle.length <= 10<sup>4</sup></code></li>
|
||||
<li><code>haystack</code> and <code>needle</code> consist of only lowercase English characters.</li>
|
||||
</ul>
|
||||
41
leetcode-cn/problem (English)/被列覆盖的最多行数(English) [maximum-rows-covered-by-columns].html
Normal file
41
leetcode-cn/problem (English)/被列覆盖的最多行数(English) [maximum-rows-covered-by-columns].html
Normal file
@ -0,0 +1,41 @@
|
||||
<p>You are given a <strong>0-indexed</strong> <code>m x n</code> binary matrix <code>mat</code> and an integer <code>cols</code>, which denotes the number of columns you must choose.</p>
|
||||
|
||||
<p>A row is <strong>covered</strong> by a set of columns if each cell in the row that has a value of <code>1</code> also lies in one of the columns of the chosen set.</p>
|
||||
|
||||
<p>Return <em>the <strong>maximum</strong> number of rows that can be <strong>covered</strong> by a set of </em><code>cols</code><em> columns.</em></p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Example 1:</strong></p>
|
||||
|
||||
<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered.png" style="width: 250px; height: 417px;" /></strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> mat = [[0,0,0],[1,0,1],[0,1,1],[0,0,1]], cols = 2
|
||||
<strong>Output:</strong> 3
|
||||
<strong>Explanation:</strong>
|
||||
As shown in the diagram above, one possible way of covering 3 rows is by selecting the 0th and 2nd columns.
|
||||
It can be shown that no more than 3 rows can be covered, so we return 3.
|
||||
</pre>
|
||||
|
||||
<p><strong>Example 2:</strong></p>
|
||||
|
||||
<p><strong><img alt="" src="https://assets.leetcode.com/uploads/2022/07/14/rowscovered2.png" style="width: 83px; height: 247px;" /></strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> mat = [[1],[0]], cols = 1
|
||||
<strong>Output:</strong> 2
|
||||
<strong>Explanation:</strong>
|
||||
Selecting the only column will result in both rows being covered, since the entire matrix is selected.
|
||||
Therefore, we return 2.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>m == mat.length</code></li>
|
||||
<li><code>n == mat[i].length</code></li>
|
||||
<li><code>1 <= m, n <= 12</code></li>
|
||||
<li><code>mat[i][j]</code> is either <code>0</code> or <code>1</code>.</li>
|
||||
<li><code>1 <= cols <= n</code></li>
|
||||
</ul>
|
||||
35
leetcode-cn/problem (English)/预算内的最多机器人数目(English) [maximum-number-of-robots-within-budget].html
Normal file
35
leetcode-cn/problem (English)/预算内的最多机器人数目(English) [maximum-number-of-robots-within-budget].html
Normal file
@ -0,0 +1,35 @@
|
||||
<p>You have <code>n</code> robots. You are given two <strong>0-indexed</strong> integer arrays, <code>chargeTimes</code> and <code>runningCosts</code>, both of length <code>n</code>. The <code>i<sup>th</sup></code> robot costs <code>chargeTimes[i]</code> units to charge and costs <code>runningCosts[i]</code> units to run. You are also given an integer <code>budget</code>.</p>
|
||||
|
||||
<p>The <strong>total cost</strong> of running <code>k</code> chosen robots is equal to <code>max(chargeTimes) + k * sum(runningCosts)</code>, where <code>max(chargeTimes)</code> is the largest charge cost among the <code>k</code> robots and <code>sum(runningCosts)</code> is the sum of running costs among the <code>k</code> robots.</p>
|
||||
|
||||
<p>Return<em> the <strong>maximum</strong> number of <strong>consecutive</strong> robots you can run such that the total cost <strong>does not</strong> exceed </em><code>budget</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> chargeTimes = [3,6,1,3,4], runningCosts = [2,1,3,4,5], budget = 25
|
||||
<strong>Output:</strong> 3
|
||||
<strong>Explanation:</strong>
|
||||
It is possible to run all individual and consecutive pairs of robots within budget.
|
||||
To obtain answer 3, consider the first 3 robots. The total cost will be max(3,6,1) + 3 * sum(2,1,3) = 6 + 3 * 6 = 24 which is less than 25.
|
||||
It can be shown that it is not possible to run more than 3 consecutive robots within budget, so we return 3.
|
||||
</pre>
|
||||
|
||||
<p><strong>Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> chargeTimes = [11,12,19], runningCosts = [10,8,7], budget = 19
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> No robot can be run that does not exceed the budget, so we return 0.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>chargeTimes.length == runningCosts.length == n</code></li>
|
||||
<li><code>1 <= n <= 5 * 10<sup>4</sup></code></li>
|
||||
<li><code>1 <= chargeTimes[i], runningCosts[i] <= 10<sup>5</sup></code></li>
|
||||
<li><code>1 <= budget <= 10<sup>15</sup></code></li>
|
||||
</ul>
|
||||
File diff suppressed because it is too large
Load Diff
File diff suppressed because one or more lines are too long
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|
||||
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"data": {
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||||
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||||
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|
||||
"questionFrontendId": "2394",
|
||||
"boundTopicId": null,
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||||
"title": "Employees With Deductions",
|
||||
"titleSlug": "employees-with-deductions",
|
||||
"content": null,
|
||||
"translatedTitle": null,
|
||||
"translatedContent": null,
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"likes": 1,
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"dislikes": 2,
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"isLiked": null,
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"similarQuestions": "[]",
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"contributors": [],
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||||
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"stats": "{\"totalAccepted\": \"86\", \"totalSubmission\": \"108\", \"totalAcceptedRaw\": 86, \"totalSubmissionRaw\": 108, \"acRate\": \"79.6%\"}",
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"hints": [],
|
||||
"solution": null,
|
||||
"status": null,
|
||||
"sampleTestCase": "{\"headers\": {\"Employees\": [\"employee_id\", \"needed_hours\"], \"Logs\": [\"employee_id\", \"in_time\", \"out_time\"]}, \"rows\": {\"Employees\": [[1, 20], [2, 12], [3, 2]], \"Logs\": [[1, \"2022-10-01 09:00:00\", \"2022-10-01 17:00:00\"],[1, \"2022-10-06 09:05:04\", \"2022-10-06 17:09:03\"], [1, \"2022-10-12 23:00:00\", \"2022-10-13 03:00:01\"], [2, \"2022-10-29 12:00:00\", \"2022-10-29 23:58:58\"]]}}",
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"metaData": "{\n \"mysql\": [\n \"Create table If Not Exists Employees (employee_id int, needed_hours int)\",\n \"Create table If Not Exists Logs (employee_id int, in_time datetime, out_time datetime)\"\n ],\n \"mssql\": [\n \"Create table Employees (employee_id int, needed_hours int)\",\n \"Create table Logs (employee_id int, in_time datetime, out_time datetime)\"\n ],\n \"oraclesql\": [\n \"Create table Employees (employee_id int, needed_hours int)\",\n \"Create table Logs (employee_id int, in_time date, out_time date)\",\n \"ALTER SESSION SET nls_date_format='YYYY-MM-DD HH24:MI:SS'\"\n ],\n \"database\": true\n}",
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"judgerAvailable": true,
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"judgeType": "large",
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"mysqlSchemas": [
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||||
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||||
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||||
"Truncate table Employees",
|
||||
"insert into Employees (employee_id, needed_hours) values ('1', '20')",
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|
||||
"insert into Employees (employee_id, needed_hours) values ('3', '2')",
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||||
"Truncate table Logs",
|
||||
"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-01 09:00:00', '2022-10-01 17:00:00')",
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||||
"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-06 09:05:04', '2022-10-06 17:09:03')",
|
||||
"insert into Logs (employee_id, in_time, out_time) values ('1', '2022-10-12 23:00:00', '2022-10-13 03:00:01')",
|
||||
"insert into Logs (employee_id, in_time, out_time) values ('2', '2022-10-29 12:00:00', '2022-10-29 23:58:58')"
|
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],
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"enableRunCode": true,
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"enableTestMode": false,
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"enableDebugger": false,
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"envInfo": "{\"mysql\": [\"MySQL\", \"<p><code>MySQL 8.0</code>.</p>\"], \"mssql\": [\"MS SQL Server\", \"<p><code>mssql server 2019</code>.</p>\"], \"oraclesql\": [\"Oracle\", \"<p><code>Oracle Sql 11.2</code>.</p>\"]}",
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File diff suppressed because one or more lines are too long
@ -0,0 +1,27 @@
|
||||
<p>Given two strings <code>needle</code> and <code>haystack</code>, return the index of the first occurrence of <code>needle</code> in <code>haystack</code>, or <code>-1</code> if <code>needle</code> is not part of <code>haystack</code>.</p>
|
||||
|
||||
<p> </p>
|
||||
<p><strong class="example">Example 1:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> haystack = "sadbutsad", needle = "sad"
|
||||
<strong>Output:</strong> 0
|
||||
<strong>Explanation:</strong> "sad" occurs at index 0 and 6.
|
||||
The first occurrence is at index 0, so we return 0.
|
||||
</pre>
|
||||
|
||||
<p><strong class="example">Example 2:</strong></p>
|
||||
|
||||
<pre>
|
||||
<strong>Input:</strong> haystack = "leetcode", needle = "leeto"
|
||||
<strong>Output:</strong> -1
|
||||
<strong>Explanation:</strong> "leeto" did not occur in "leetcode", so we return -1.
|
||||
</pre>
|
||||
|
||||
<p> </p>
|
||||
<p><strong>Constraints:</strong></p>
|
||||
|
||||
<ul>
|
||||
<li><code>1 <= haystack.length, needle.length <= 10<sup>4</sup></code></li>
|
||||
<li><code>haystack</code> and <code>needle</code> consist of only lowercase English characters.</li>
|
||||
</ul>
|
||||
Loading…
x
Reference in New Issue
Block a user