mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
synced 2025-01-10 18:48:13 +08:00
189 lines
24 KiB
JSON
189 lines
24 KiB
JSON
{
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"questionId": "2006",
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"questionFrontendId": "1894",
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"categoryTitle": "Algorithms",
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"boundTopicId": 820492,
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"title": "Find the Student that Will Replace the Chalk",
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"titleSlug": "find-the-student-that-will-replace-the-chalk",
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"content": "<p>There are <code>n</code> students in a class numbered from <code>0</code> to <code>n - 1</code>. The teacher will give each student a problem starting with the student number <code>0</code>, then the student number <code>1</code>, and so on until the teacher reaches the student number <code>n - 1</code>. After that, the teacher will restart the process, starting with the student number <code>0</code> again.</p>\n\n<p>You are given a <strong>0-indexed</strong> integer array <code>chalk</code> and an integer <code>k</code>. There are initially <code>k</code> pieces of chalk. When the student number <code>i</code> is given a problem to solve, they will use <code>chalk[i]</code> pieces of chalk to solve that problem. However, if the current number of chalk pieces is <strong>strictly less</strong> than <code>chalk[i]</code>, then the student number <code>i</code> will be asked to <strong>replace</strong> the chalk.</p>\n\n<p>Return <em>the <strong>index</strong> of the student that will <strong>replace</strong> the chalk pieces</em>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> chalk = [5,1,5], k = 22\n<strong>Output:</strong> 0\n<strong>Explanation: </strong>The students go in turns as follows:\n- Student number 0 uses 5 chalk, so k = 17.\n- Student number 1 uses 1 chalk, so k = 16.\n- Student number 2 uses 5 chalk, so k = 11.\n- Student number 0 uses 5 chalk, so k = 6.\n- Student number 1 uses 1 chalk, so k = 5.\n- Student number 2 uses 5 chalk, so k = 0.\nStudent number 0 does not have enough chalk, so they will have to replace it.</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> chalk = [3,4,1,2], k = 25\n<strong>Output:</strong> 1\n<strong>Explanation: </strong>The students go in turns as follows:\n- Student number 0 uses 3 chalk so k = 22.\n- Student number 1 uses 4 chalk so k = 18.\n- Student number 2 uses 1 chalk so k = 17.\n- Student number 3 uses 2 chalk so k = 15.\n- Student number 0 uses 3 chalk so k = 12.\n- Student number 1 uses 4 chalk so k = 8.\n- Student number 2 uses 1 chalk so k = 7.\n- Student number 3 uses 2 chalk so k = 5.\n- Student number 0 uses 3 chalk so k = 2.\nStudent number 1 does not have enough chalk, so they will have to replace it.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>chalk.length == n</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= chalk[i] <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= k <= 10<sup>9</sup></code></li>\n</ul>\n",
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"translatedTitle": "找到需要补充粉笔的学生编号",
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"translatedContent": "<p>一个班级里有 <code>n</code> 个学生,编号为 <code>0</code> 到 <code>n - 1</code> 。每个学生会依次回答问题,编号为 <code>0</code> 的学生先回答,然后是编号为 <code>1</code> 的学生,以此类推,直到编号为 <code>n - 1</code> 的学生,然后老师会重复这个过程,重新从编号为 <code>0</code> 的学生开始回答问题。</p>\n\n<p>给你一个长度为 <code>n</code> 且下标从 <code>0</code> 开始的整数数组 <code>chalk</code> 和一个整数 <code>k</code> 。一开始粉笔盒里总共有 <code>k</code> 支粉笔。当编号为 <code>i</code> 的学生回答问题时,他会消耗 <code>chalk[i]</code> 支粉笔。如果剩余粉笔数量 <strong>严格小于</strong> <code>chalk[i]</code> ,那么学生 <code>i</code> 需要 <strong>补充</strong> 粉笔。</p>\n\n<p>请你返回需要 <strong>补充</strong> 粉笔的学生 <strong>编号</strong> 。</p>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1:</strong></p>\n\n<pre>\n<b>输入:</b>chalk = [5,1,5], k = 22\n<b>输出:</b>0\n<strong>解释:</strong>学生消耗粉笔情况如下:\n- 编号为 0 的学生使用 5 支粉笔,然后 k = 17 。\n- 编号为 1 的学生使用 1 支粉笔,然后 k = 16 。\n- 编号为 2 的学生使用 5 支粉笔,然后 k = 11 。\n- 编号为 0 的学生使用 5 支粉笔,然后 k = 6 。\n- 编号为 1 的学生使用 1 支粉笔,然后 k = 5 。\n- 编号为 2 的学生使用 5 支粉笔,然后 k = 0 。\n编号为 0 的学生没有足够的粉笔,所以他需要补充粉笔。</pre>\n\n<p><strong class=\"example\">示例 2:</strong></p>\n\n<pre>\n<b>输入:</b>chalk = [3,4,1,2], k = 25\n<b>输出:</b>1\n<b>解释:</b>学生消耗粉笔情况如下:\n- 编号为 0 的学生使用 3 支粉笔,然后 k = 22 。\n- 编号为 1 的学生使用 4 支粉笔,然后 k = 18 。\n- 编号为 2 的学生使用 1 支粉笔,然后 k = 17 。\n- 编号为 3 的学生使用 2 支粉笔,然后 k = 15 。\n- 编号为 0 的学生使用 3 支粉笔,然后 k = 12 。\n- 编号为 1 的学生使用 4 支粉笔,然后 k = 8 。\n- 编号为 2 的学生使用 1 支粉笔,然后 k = 7 。\n- 编号为 3 的学生使用 2 支粉笔,然后 k = 5 。\n- 编号为 0 的学生使用 3 支粉笔,然后 k = 2 。\n编号为 1 的学生没有足够的粉笔,所以他需要补充粉笔。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>chalk.length == n</code></li>\n\t<li><code>1 <= n <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= chalk[i] <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= k <= 10<sup>9</sup></code></li>\n</ul>\n",
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"Subtract the sum of chalk from k until k is less than the sum of chalk.",
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"Now iterate over the array. If chalk[i] is less than k, this is the answer. Otherwise, subtract chalk[i] from k and continue."
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