Package g2801_2900.s2842_count_k_subsequences_of_a_string_with_maximum_beauty

Class Solution

java.lang.Object
g2801_2900.s2842_count_k_subsequences_of_a_string_with_maximum_beauty.Solution
public class Solution extends Object
2842 - Count K-Subsequences of a String With Maximum Beauty.

Hard

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 s are:
    • “ ab bbdd” -> "ab" having a beauty of f('a') + f('b') = 4
    • “ a bbb**d**d” -> "ad" having a beauty of f('a') + f('d') = 3
    • “a**b**bb d d” -> "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 c occurs 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:

**bc**ca having a beauty of f(‘b’) + f(‘c’) = 3

**b**c c a having a beauty of f(‘b’) + f(‘c’) = 3

bcca having a beauty of f(‘b’) + f(‘a’) = 2

b**c**c a having a beauty of f(‘c’) + f(‘a’) = 3

bc**ca** 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:

ab b**cd** having a beauty of f(‘a’) + f(‘b’) + f(‘c’) + f(‘d’) = 5

a b bcd 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
  • s consists only of lowercase English letters.

  • Constructor Details

    • Solution

      public Solution()

  • Method Details

    • countKSubsequencesWithMaxBeauty

      public int countKSubsequencesWithMaxBeauty(String s, int k)