Count K-Subsequences of a String With Maximum Beauty

Count K-Subsequences of a String With Maximum Beauty - Problem

Imagine you're a talent scout looking for the most beautiful teams of unique performers from a group of artists. Each artist type appears multiple times, and a team's beauty is determined by how popular each artist type is!

Given a string s and an integer k, you need to find all possible k-subsequences with maximum beauty.

What's a k-subsequence?
A subsequence of length k where all characters are unique (no duplicates allowed).

What's beauty?
For each character c in the subsequence, let f(c) be how many times c appears in the original string s. The beauty is the sum of all f(c) values.

Example: s = "abbbdd", k = 2
• f('a') = 1, f('b') = 3, f('d') = 2
• Subsequence "ab" has beauty = 1 + 3 = 4
• Subsequence "ad" has beauty = 1 + 2 = 3
• Subsequence "bd" has beauty = 3 + 2 = 5 ← Maximum beauty!

Goal: Count how many k-subsequences achieve the maximum possible beauty. Return the answer modulo 10⁹ + 7.

Input & Output

example_1.py — Basic case

$ Input: s = "abbbdd", k = 2

› Output: 1

💡 Note: Character frequencies: f('a')=1, f('b')=3, f('d')=2. Possible 2-subsequences: 'ab' (beauty=4), 'ad' (beauty=3), 'bd' (beauty=5). Maximum beauty is 5, achieved by 1 subsequence.

example_2.py — Multiple optimal solutions

$ Input: s = "aabbcc", k = 2

› Output: 3

💡 Note: All characters have frequency 2. All 2-subsequences ('ab', 'ac', 'bc') have beauty 4. So 3 subsequences achieve maximum beauty.

example_3.py — Insufficient unique characters

$ Input: s = "aaa", k = 2

› Output: 0

💡 Note: Only 1 unique character 'a', but we need k=2 unique characters. No valid k-subsequence exists.

Visualization

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Understanding the Visualization

Talent Assessment

Count how popular each performer type is (character frequency)

Ranking

Rank all performer types by their popularity in descending order

Greedy Selection

Always pick the top k most popular types for maximum total talent

Handle Ties

When performers have equal popularity, use combinatorics to count selection ways

Key Takeaway

🎯 Key Insight: The optimal k-subsequence always contains the k most frequent characters. Use combinatorics to count arrangements when frequencies tie, achieving O(n + m log m) complexity instead of exponential brute force.

Time & Space Complexity

Time Complexity

⏱️

O(n + m log m)

O(n) to count frequencies, O(m log m) to sort m unique characters

⚡ Linearithmic

Space Complexity

O(m)

O(m) space for frequency map where m is number of unique characters

✓ Linear Space

Constraints

1 ≤ s.length ≤ 2 × 10⁴
1 ≤ k ≤ s.length
s consists of only lowercase English letters
Answer fits in a 32-bit integer after taking modulo

Asked in

G Google 45 f Meta 38 a Amazon 32 ⊞ Microsoft 28

This problem requires finding k-subsequences with maximum beauty (sum of character frequencies). The key insight is that optimal subsequences always contain the k most frequent characters. Use greedy selection with combinatorics to handle ties efficiently, avoiding exponential enumeration.

Common Approaches

Approach	Time	Space	Notes
✓ Greedy + Combinatorics (Optimal)	O(n + m log m)	O(m)	Use greedy selection of most frequent characters with combinatorial counting
Brute Force (Generate All Subsequences)	O(C(n,k) * k)	O(n)	Generate all possible k-subsequences and find maximum beauty

Greedy + Combinatorics (Optimal) — Algorithm Steps

Count frequency of each character
Sort characters by frequency in descending order
Greedily select the k most frequent characters
Handle ties using combinatorics - if some characters have the same frequency, calculate how many ways to choose among them
Use modular arithmetic for the final answer

Visualization

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Step-by-Step Walkthrough

Count & Sort

Count character frequencies and sort in descending order

Greedy Selection

Select the k most frequent characters for maximum beauty

Handle Ties

Use combinatorics when multiple characters have the same frequency

Calculate Ways

Compute nCr for the number of valid selections

Code -

solution.c — C

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

#define MOD 1000000007
#define MAXN 26

long long modPow(long long base, long long exp, long long mod) {
    long long result = 1;
    base %= mod;
    while (exp > 0) {
        if (exp & 1) {
            result = (result * base) % mod;
        }
        base = (base * base) % mod;
        exp >>= 1;
    }
    return result;
}

long long modInv(long long a, long long mod) {
    return modPow(a, mod - 2, mod);
}

long long nCr(int n, int r) {
    if (r > n || r < 0) return 0;
    if (r == 0 || r == n) return 1;
    
    long long num = 1, den = 1;
    for (int i = 0; i < r; i++) {
        num = (num * (n - i)) % MOD;
        den = (den * (i + 1)) % MOD;
    }
    
    return (num * modInv(den, MOD)) % MOD;
}

int compare(const void *a, const void *b) {
    return (*(int*)b - *(int*)a);
}

int countKSubsequencesWithMaxBeauty(char* s, int k) {
    int freq[MAXN] = {0};
    int len = strlen(s);
    
    for (int i = 0; i < len; i++) {
        freq[s[i] - 'a']++;
    }
    
    int freqValues[MAXN];
    int count = 0;
    
    for (int i = 0; i < MAXN; i++) {
        if (freq[i] > 0) {
            freqValues[count++] = freq[i];
        }
    }
    
    if (count < k) return 0;
    
    qsort(freqValues, count, sizeof(int), compare);
    
    int kthFreq = freqValues[k - 1];
    
    int mustInclude = 0;
    for (int i = 0; i < count; i++) {
        if (freqValues[i] > kthFreq) {
            mustInclude++;
        } else {
            break;
        }
    }
    
    int canInclude = 0;
    for (int i = 0; i < count; i++) {
        if (freqValues[i] == kthFreq) {
            canInclude++;
        }
    }
    
    int needToChoose = k - mustInclude;
    
    if (needToChoose > canInclude) return 0;
    
    return (int)nCr(canInclude, needToChoose);
}

int main() {
    char s[100000];
    int k;
    scanf("%s", s);
    
    int len = strlen(s);
    if (s[0] == '"' && s[len-1] == '"') {
        for (int i = 1; i < len-1; i++) {
            s[i-1] = s[i];
        }
        s[len-2] = '\0';
    }
    
    scanf("%d", &k);
    printf("%d\n", countKSubsequencesWithMaxBeauty(s, k));
    return 0;
}

Time & Space Complexity

Time Complexity

⏱️

O(n + m log m)

O(n) to count frequencies, O(m log m) to sort m unique characters

⚡ Linearithmic

Space Complexity

O(m)

O(m) space for frequency map where m is number of unique characters

✓ Linear Space

Constraints

1 ≤ s.length ≤ 2 × 10⁴
1 ≤ k ≤ s.length
s consists of only lowercase English letters
Answer fits in a 32-bit integer after taking modulo

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Visualization

Time & Space Complexity

Related Problems

Constraints

Common Approaches

Greedy + Combinatorics (Optimal) — Algorithm Steps

Visualization

Code -

Time & Space Complexity

Constraints

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