You are given a string s and an integer k.
A k-subsequence is a subsequence of s, having length k, and all its characters are unique, i.e., every character occurs once.
Let f(c) denote the number of times the character c occurs in s.
The beauty of a k-subsequence is the sum of f(c) for every character c in the k-subsequence.
For example, consider s = "abbbdd" and k = 2:
f('a') = 1, f('b') = 3, f('d') = 2s are:\n\t"abbbdd" -> "ab" having a beauty of f('a') + f('b') = 4"abbbdd" -> "ad" having a beauty of f('a') + f('d') = 3"abbbdd" -> "bd" having a beauty of f('b') + f('d') = 5Return an integer denoting the number of k-subsequences whose beauty is the maximum among all k-subsequences. Since the answer may be too large, return it modulo 109 + 7.
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.
\n\nNotes
\n\nf(c) is the number of times a character c occurs in s, not a k-subsequence.\n
Example 1:
\n\n\nInput: s = "bcca", k = 2\nOutput: 4\nExplanation: From s we have f('a') = 1, f('b') = 1, and f('c') = 2.\nThe k-subsequences of s are: \nbcca having a beauty of f('b') + f('c') = 3 \nbcca having a beauty of f('b') + f('c') = 3 \nbcca having a beauty of f('b') + f('a') = 2 \nbcca having a beauty of f('c') + f('a') = 3\nbcca 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\n\nExample 2:
\n\n\nInput: s = "abbcd", k = 4\nOutput: 2\nExplanation: From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. \nThe k-subsequences of s are: \nabbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5\nabbcd 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\n\n\n
Constraints:
\n\n1 <= s.length <= 2 * 1051 <= k <= s.lengths consists only of lowercase English letters.给你一个字符串 s 和一个整数 k 。
k 子序列指的是 s 的一个长度为 k 的 子序列 ，且所有字符都是 唯一 的，也就是说每个字符在子序列里只出现过一次。
定义 f(c) 为字符 c 在 s 中出现的次数。
k 子序列的 美丽值 定义为这个子序列中每一个字符 c 的 f(c) 之 和 。
比方说，s = \"abbbdd\" 和 k = 2 ，我们有：
f('a') = 1, f('b') = 3, f('d') = 2s 的部分 k 子序列为：\n\t\"abbbdd\" -> \"ab\" ，美丽值为 f('a') + f('b') = 4\"abbbdd\" -> \"ad\" ，美丽值为 f('a') + f('d') = 3\"abbbdd\" -> \"bd\" ，美丽值为 f('b') + f('d') = 5请你返回一个整数，表示所有 k 子序列 里面 美丽值 是 最大值 的子序列数目。由于答案可能很大，将结果对 109 + 7 取余后返回。
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。
\n\n注意：
\n\nf(c) 指的是字符 c 在字符串 s 的出现次数，不是在 k 子序列里的出现次数。\n\n
示例 1：
\n\n\n输入：s = \"bcca\", k = 2\n输出：4\n解释：s 中我们有 f('a') = 1 ，f('b') = 1 和 f('c') = 2 。\ns 的 k 子序列为：\nbcca ，美丽值为 f('b') + f('c') = 3\nbcca ，美丽值为 f('b') + f('c') = 3\nbcca ，美丽值为 f('b') + f('a') = 2\nbcca ，美丽值为 f('c') + f('a') = 3\nbcca ，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\n\n示例 2：
\n\n\n输入：s = \"abbcd\", k = 4\n输出：2\n解释：s 中我们有 f('a') = 1 ，f('b') = 2 ，f('c') = 1 和 f('d') = 1 。\ns 的 k 子序列为：\nabbcd ，美丽值为 f('a') + f('b') + f('c') + f('d') = 5\nabbcd ，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\n\n
\n\n
提示：
\n\n1 <= s.length <= 2 * 1051 <= k <= s.lengths 只包含小写英文字母。k characters such that the sum of their frequencies is maximum.",
"An obvious case to eliminate is if k is greater than the number of distinct characters in s, then the answer is 0.",
"We are now interested in the top frequencies among the characters. Using a map data structure, let cnt[x] denote the number of characters that have a frequency of x.",
"Starting from the maximum value x in cnt. Let i = min(k, cnt[x]) we add to our result cnt[x]Ci * xi representing the number of ways to select i characters from all characters with frequency x, multiplied by the number of ways to choose each individual character. Subtract i from k and continue downwards to the next maximum value.",
"Powers, combinations, and additions should be done modulo 109 + 7."
],
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"status": null,
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