Is Count K-Subsequences of a String With Maximum Beauty easy or hard?

The problem is rated Hard because it combines multiple concepts: greedy optimization, frequency counting, and combinatorial math. Many candidates identify the greedy selection but struggle with counting valid subsequences when frequencies tie. Handling these cases correctly with combinations is the key challenge.

Count K-Subsequences of a String With Maximum Beauty Python/Java solution

In Python, use collections.Counter to count character frequencies, sort them, and multiply the selected frequencies while handling ties with combinatorics. In Java, use a HashMap or int[26] for counting, sort the values, and compute combinations with modular arithmetic. Both implementations follow the same greedy + combinatorial logic.

How to solve Count K-Subsequences of a String With Maximum Beauty in O(n)?

Pure O(n) is difficult because the algorithm usually requires sorting character frequencies. However, since the alphabet size is small, the effective complexity becomes O(n + 26 log 26), which behaves like linear time in practice. The process involves counting frequencies, selecting the largest k values greedily, and computing combinations for ties.

What is the best approach for Count K-Subsequences of a String With Maximum Beauty?

The most effective solution uses a greedy strategy with frequency counting and combinatorics. Count character frequencies, sort them in descending order, and select the top k frequencies to maximize beauty. When multiple characters share the same cutoff frequency, use combinations to count valid selections. This runs in O(n + m log m) time where m is the number of unique characters.

Is Count K-Subsequences of a String With Maximum Beauty asked at Google/Amazon/Meta?

Problems combining greedy selection with combinatorics frequently appear in interviews at companies like Google, Amazon, and Meta. Variants that involve selecting optimal characters, counting subsequences, or handling frequency ties are common in advanced algorithm rounds.

What data structure is used in Count K-Subsequences of a String With Maximum Beauty?

The main data structure is a hash table or frequency array to count occurrences of each character. After counting, a sorted array or priority ordering of frequencies is used to apply the greedy selection. Combinatorial calculations handle cases where multiple characters share the same frequency.

What is the time complexity of Count K-Subsequences of a String With Maximum Beauty?

The optimal solution runs in O(n + m log m) time and O(m) space. Counting frequencies takes O(n), sorting unique character frequencies takes O(m log m), and combinatorial calculations are constant or logarithmic depending on implementation. Since the alphabet size is typically small (like 26 for lowercase letters), the sorting cost is minimal.

Description Hints 5 Solution Video 25:57

2842. Count K-Subsequences of a String With Maximum Beauty

Hard30.2% acceptance1 company

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:
- "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 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: 
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
s consists only of lowercase English letters.

Asked at 1 company

Google

Topics

Hints open one at a time

Hint 1

Since every character appears once in a k-subsequence, we can solve the following problem first: Find the total number of ways to select k characters such that the sum of their frequencies is maximum.

Hint 2

An obvious case to eliminate is if k is greater than the number of distinct characters in s, then the answer is 0.

Hint 3

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.

Hint 4

Starting from the maximum value x in cnt. Let i = min(k, cnt[x]) we add to our result ^cnt[x]C_i * 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.

Hint 5

Powers, combinations, and additions should be done modulo 10⁹ + 7.

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