mirror of
https://gitee.com/coder-xiaomo/leetcode-problemset
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183 lines
24 KiB
JSON
183 lines
24 KiB
JSON
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"categoryTitle": "Algorithms",
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"title": "Find Missing Observations",
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"content": "<p>You have observations of <code>n + m</code> <strong>6-sided</strong> dice rolls with each face numbered from <code>1</code> to <code>6</code>. <code>n</code> of the observations went missing, and you only have the observations of <code>m</code> rolls. Fortunately, you have also calculated the <strong>average value</strong> of the <code>n + m</code> rolls.</p>\n\n<p>You are given an integer array <code>rolls</code> of length <code>m</code> where <code>rolls[i]</code> is the value of the <code>i<sup>th</sup></code> observation. You are also given the two integers <code>mean</code> and <code>n</code>.</p>\n\n<p>Return <em>an array of length </em><code>n</code><em> containing the missing observations such that the <strong>average value </strong>of the </em><code>n + m</code><em> rolls is <strong>exactly</strong> </em><code>mean</code>. If there are multiple valid answers, return <em>any of them</em>. If no such array exists, return <em>an empty array</em>.</p>\n\n<p>The <strong>average value</strong> of a set of <code>k</code> numbers is the sum of the numbers divided by <code>k</code>.</p>\n\n<p>Note that <code>mean</code> is an integer, so the sum of the <code>n + m</code> rolls should be divisible by <code>n + m</code>.</p>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> rolls = [3,2,4,3], mean = 4, n = 2\n<strong>Output:</strong> [6,6]\n<strong>Explanation:</strong> The mean of all n + m rolls is (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4.\n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> rolls = [1,5,6], mean = 3, n = 4\n<strong>Output:</strong> [2,3,2,2]\n<strong>Explanation:</strong> The mean of all n + m rolls is (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3.\n</pre>\n\n<p><strong class=\"example\">Example 3:</strong></p>\n\n<pre>\n<strong>Input:</strong> rolls = [1,2,3,4], mean = 6, n = 4\n<strong>Output:</strong> []\n<strong>Explanation:</strong> It is impossible for the mean to be 6 no matter what the 4 missing rolls are.\n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>m == rolls.length</code></li>\n\t<li><code>1 <= n, m <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= rolls[i], mean <= 6</code></li>\n</ul>\n",
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"translatedTitle": "找出缺失的观测数据",
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"translatedContent": "<p>现有一份 <code>n + m</code> 次投掷单个<strong> 六面</strong> 骰子的观测数据,骰子的每个面从 <code>1</code> 到 <code>6</code> 编号。观测数据中缺失了 <code>n</code> 份,你手上只拿到剩余 <code>m</code> 次投掷的数据。幸好你有之前计算过的这 <code>n + m</code> 次投掷数据的 <strong>平均值</strong> 。</p>\n\n<p>给你一个长度为 <code>m</code> 的整数数组 <code>rolls</code> ,其中 <code>rolls[i]</code> 是第 <code>i</code> 次观测的值。同时给你两个整数 <code>mean</code> 和 <code>n</code> 。</p>\n\n<p>返回一个长度为<em> </em><code>n</code><em> </em>的数组,包含所有缺失的观测数据,且满足这<em> </em><code>n + m</code><em> </em>次投掷的 <strong>平均值</strong> 是<em> </em><code>mean</code> 。如果存在多组符合要求的答案,只需要返回其中任意一组即可。如果不存在答案,返回一个空数组。</p>\n\n<p><code>k</code> 个数字的 <strong>平均值</strong> 为这些数字求和后再除以 <code>k</code> 。</p>\n\n<p>注意 <code>mean</code> 是一个整数,所以 <code>n + m</code> 次投掷的总和需要被 <code>n + m</code> 整除。</p>\n\n<p> </p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong>rolls = [3,2,4,3], mean = 4, n = 2\n<strong>输出:</strong>[6,6]\n<strong>解释:</strong>所有 n + m 次投掷的平均值是 (3 + 2 + 4 + 3 + 6 + 6) / 6 = 4 。\n</pre>\n\n<p><strong>示例 2:</strong></p>\n\n<pre>\n<strong>输入:</strong>rolls = [1,5,6], mean = 3, n = 4\n<strong>输出:</strong>[2,3,2,2]\n<strong>解释:</strong>所有 n + m 次投掷的平均值是 (1 + 5 + 6 + 2 + 3 + 2 + 2) / 7 = 3 。\n</pre>\n\n<p><strong>示例 3:</strong></p>\n\n<pre>\n<strong>输入:</strong>rolls = [1,2,3,4], mean = 6, n = 4\n<strong>输出:</strong>[]\n<strong>解释:</strong>无论丢失的 4 次数据是什么,平均值都不可能是 6 。\n</pre>\n\n<p><strong>示例 4:</strong></p>\n\n<pre>\n<strong>输入:</strong>rolls = [1], mean = 3, n = 1\n<strong>输出:</strong>[5]\n<strong>解释:</strong>所有 n + m 次投掷的平均值是 (1 + 5) / 2 = 3 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示:</strong></p>\n\n<ul>\n\t<li><code>m == rolls.length</code></li>\n\t<li><code>1 <= n, m <= 10<sup>5</sup></code></li>\n\t<li><code>1 <= rolls[i], mean <= 6</code></li>\n</ul>\n",
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