- Difficulty: Hard
- Tags: LeetCode, Hard, Greedy, Hash Table, Math, String, Combinatorics, leetcode-2842, O(n), O(1), Quick Select
Problem
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') = 2
- Some k-subsequences of sare:- "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') = 5
 
Return 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.
Notes
- f(c)is the number of times a character- coccurs in- s, not a k-subsequence.
- 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.
Example 1:
Input: s = "bcca", k = 2
Output: 4
Explanation: From s we have f('a') = 1, f('b') = 1, and f('c') = 2.
The k-subsequences of s are: 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('c') = 3 
bcca having a beauty of f('b') + f('a') = 2 
bcca having a beauty of f('c') + f('a') = 3
bcca having a beauty of f('c') + f('a') = 3 
There are 4 k-subsequences that have the maximum beauty, 3. 
Hence, the answer is 4. 
Example 2:
Input: s = "abbcd", k = 4
Output: 2
Explanation: From s we have f('a') = 1, f('b') = 2, f('c') = 1, and f('d') = 1. 
The k-subsequences of s are: 
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5
abbcd having a beauty of f('a') + f('b') + f('c') + f('d') = 5 
There are 2 k-subsequences that have the maximum beauty, 5. 
Hence, the answer is 2. 
Constraints:
- 1 <= s.length <= 2 * 105
- 1 <= k <= s.length
- sconsists only of lowercase English letters.