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https://gitee.com/coder-xiaomo/leetcode-problemset
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198 lines
29 KiB
JSON
198 lines
29 KiB
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"title": "Count K-Subsequences of a String With Maximum Beauty",
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"content": "<p>You are given a string <code>s</code> and an integer <code>k</code>.</p>\n\n<p>A <strong>k-subsequence</strong> is a <strong>subsequence</strong> of <code>s</code>, having length <code>k</code>, and all its characters are <strong>unique</strong>, <strong>i.e</strong>., every character occurs once.</p>\n\n<p>Let <code>f(c)</code> denote the number of times the character <code>c</code> occurs in <code>s</code>.</p>\n\n<p>The <strong>beauty</strong> of a <strong>k-subsequence</strong> is the <strong>sum</strong> of <code>f(c)</code> for every character <code>c</code> in the k-subsequence.</p>\n\n<p>For example, consider <code>s = "abbbdd"</code> and <code>k = 2</code>:</p>\n\n<ul>\n\t<li><code>f('a') = 1</code>, <code>f('b') = 3</code>, <code>f('d') = 2</code></li>\n\t<li>Some k-subsequences of <code>s</code> are:\n\t<ul>\n\t\t<li><code>"<u><strong>ab</strong></u>bbdd"</code> -> <code>"ab"</code> having a beauty of <code>f('a') + f('b') = 4</code></li>\n\t\t<li><code>"<u><strong>a</strong></u>bbb<strong><u>d</u></strong>d"</code> -> <code>"ad"</code> having a beauty of <code>f('a') + f('d') = 3</code></li>\n\t\t<li><code>"a<strong><u>b</u></strong>bb<u><strong>d</strong></u>d"</code> -> <code>"bd"</code> having a beauty of <code>f('b') + f('d') = 5</code></li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>Return <em>an integer denoting the number of k-subsequences </em><em>whose <strong>beauty</strong> is the <strong>maximum</strong> among all <strong>k-subsequences</strong></em>. Since the answer may be too large, return it modulo <code>10<sup>9</sup> + 7</code>.</p>\n\n<p>A subsequence of a string is a new string formed from the original string by deleting some (possibly none) of the characters without disturbing the relative positions of the remaining characters.</p>\n\n<p><strong>Notes</strong></p>\n\n<ul>\n\t<li><code>f(c)</code> is the number of times a character <code>c</code> occurs in <code>s</code>, not a k-subsequence.</li>\n\t<li>Two k-subsequences are considered different if one is formed by an index that is not present in the other. So, two k-subsequences may form the same string.</li>\n</ul>\n\n<p> </p>\n<p><strong class=\"example\">Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "bcca", k = 2\n<strong>Output:</strong> 4\n<strong>Explanation:</strong> <span style=\"white-space: normal\">From s we have f('a') = 1, f('b') = 1, and f('c') = 2.</span>\nThe k-subsequences of s are: \n<strong><u>bc</u></strong>ca having a beauty of f('b') + f('c') = 3 \n<strong><u>b</u></strong>c<u><strong>c</strong></u>a having a beauty of f('b') + f('c') = 3 \n<strong><u>b</u></strong>cc<strong><u>a</u></strong> having a beauty of f('b') + f('a') = 2 \nb<strong><u>c</u></strong>c<u><strong>a</strong></u><strong> </strong>having a beauty of f('c') + f('a') = 3\nbc<strong><u>ca</u></strong> having a beauty of f('c') + f('a') = 3 \nThere are 4 k-subsequences that have the maximum beauty, 3. \nHence, the answer is 4. \n</pre>\n\n<p><strong class=\"example\">Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> s = "abbcd", k = 4\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. \nThe k-subsequences of s are: \n<u><strong>ab</strong></u>b<strong><u>cd</u></strong> having a beauty of f('a') + f('b') + f('c') + f('d') = 5\n<u style=\"white-space: normal;\"><strong>a</strong></u>b<u><strong>bcd</strong></u> having a beauty of f('a') + f('b') + f('c') + f('d') = 5 \nThere are 2 k-subsequences that have the maximum beauty, 5. \nHence, the answer is 2. \n</pre>\n\n<p> </p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length <= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 <= k <= s.length</code></li>\n\t<li><code>s</code> consists only of lowercase English letters.</li>\n</ul>\n",
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"translatedTitle": "统计一个字符串的 k 子序列美丽值最大的数目",
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"translatedContent": "<p>给你一个字符串 <code>s</code> 和一个整数 <code>k</code> 。</p>\n\n<p><strong>k 子序列</strong>指的是 <code>s</code> 的一个长度为 <code>k</code> 的 <strong>子序列</strong> ，且所有字符都是 <strong>唯一</strong> 的，也就是说每个字符在子序列里只出现过一次。</p>\n\n<p>定义 <code>f(c)</code> 为字符 <code>c</code> 在 <code>s</code> 中出现的次数。</p>\n\n<p>k 子序列的 <strong>美丽值</strong> 定义为这个子序列中每一个字符 <code>c</code> 的 <code>f(c)</code> 之 <strong>和</strong> 。</p>\n\n<p>比方说，<code>s = \"abbbdd\"</code> 和 <code>k = 2</code> ，我们有：</p>\n\n<ul>\n\t<li><code>f('a') = 1</code>, <code>f('b') = 3</code>, <code>f('d') = 2</code></li>\n\t<li><code>s</code> 的部分 k 子序列为：\n\t<ul>\n\t\t<li><code>\"<em><strong>ab</strong></em>bbdd\"</code> -> <code>\"ab\"</code> ，美丽值为 <code>f('a') + f('b') = 4</code></li>\n\t\t<li><code>\"<em><strong>a</strong></em>bbb<em><strong>d</strong></em>d\"</code> -> <code>\"ad\"</code> ，美丽值为 <code>f('a') + f('d') = 3</code></li>\n\t\t<li><code>\"a<em><strong>b</strong></em>bb<em><strong>d</strong></em>d\"</code> -> <code>\"bd\"</code> ，美丽值为 <code>f('b') + f('d') = 5</code></li>\n\t</ul>\n\t</li>\n</ul>\n\n<p>请你返回一个整数，表示所有 <strong>k 子序列 </strong>里面 <strong>美丽值 </strong>是 <strong>最大值</strong> 的子序列数目。由于答案可能很大，将结果对 <code>10<sup>9</sup> + 7</code> 取余后返回。</p>\n\n<p>一个字符串的子序列指的是从原字符串里面删除一些字符（也可能一个字符也不删除），不改变剩下字符顺序连接得到的新字符串。</p>\n\n<p><strong>注意：</strong></p>\n\n<ul>\n\t<li><code>f(c)</code> 指的是字符 <code>c</code> 在字符串 <code>s</code> 的出现次数，不是在 k 子序列里的出现次数。</li>\n\t<li>两个 k 子序列如果有任何一个字符在原字符串中的下标不同，则它们是两个不同的子序列。所以两个不同的 k 子序列可能产生相同的字符串。</li>\n</ul>\n\n<p> </p>\n\n<p><strong class=\"example\">示例 1：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"bcca\", k = 2\n<b>输出：</b>4\n<b>解释：</b><span style=\"white-space: normal\">s 中我们有 f('a') = 1 ，f('b') = 1 和 f('c') = 2 。</span>\ns 的 k 子序列为：\n<em><strong>bc</strong></em>ca ，美丽值为 f('b') + f('c') = 3\n<em><strong>b</strong></em>c<em><strong>c</strong></em>a ，美丽值为 f('b') + f('c') = 3\n<em><strong>b</strong></em>cc<em><strong>a</strong></em> ，美丽值为 f('b') + f('a') = 2\nb<em><strong>c</strong></em>c<em><strong>a</strong></em><strong> </strong>，美丽值为 f('c') + f('a') = 3\nbc<em><strong>ca</strong></em> ，美丽值为 f('c') + f('a') = 3\n总共有 4 个 k 子序列美丽值为最大值 3 。\n所以答案为 4 。\n</pre>\n\n<p><strong class=\"example\">示例 2：</strong></p>\n\n<pre>\n<b>输入：</b>s = \"abbcd\", k = 4\n<b>输出：</b>2\n<b>解释：</b><span style=\"white-space: normal\">s 中我们有 f('a') = 1 ，f('b') = 2 ，f('c') = 1 和</span> f('d') = 1 。\ns 的 k 子序列为：\n<em><strong>ab</strong></em>b<em><strong>cd</strong></em> ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5\n<span style=\"white-space: normal;\"><b><i>a</i></b></span>b<em><strong>bcd</strong></em> ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5 \n总共有 2 个 k 子序列美丽值为最大值 5 。\n所以答案为 2 。\n</pre>\n\n<p> </p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>1 <= s.length <= 2 * 10<sup>5</sup></code></li>\n\t<li><code>1 <= k <= s.length</code></li>\n\t<li><code>s</code> 只包含小写英文字母。</li>\n</ul>\n",
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"Since every character appears once in a k-subsequence, we can solve the following problem first: Find the total number of ways to select <code>k</code> characters such that the sum of their frequencies is maximum.",
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"An obvious case to eliminate is if <code>k</code> is greater than the number of distinct characters in <code>s</code>, then the answer is <code>0</code>.",
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"We are now interested in the top frequencies among the characters. Using a map data structure, let <code>cnt[x]</code> denote the number of characters that have a frequency of <code>x</code>.",
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"Starting from the maximum value <code>x</code> in <code>cnt</code>. Let <code>i = min(k, cnt[x])</code> we add to our result <code> <sup>cnt[x]</sup>C<sub>i</sub> * x<sup>i</sup></code> representing the number of ways to select <code>i</code> characters from all characters with frequency <code>x</code>, multiplied by the number of ways to choose each individual character. Subtract <code>i</code> from <code>k</code> and continue downwards to the next maximum value.",
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