Introduction

In this blog post, we will explore how to solve the LeetCode problem "2842" using AI. LeetCode is a popular platform for preparing for coding interviews, and with the help of AI tools like Chatmagic, we can generate solutions quickly and efficiently - helping you pass the interviews and get the job offer without having to study for months.

Problem Statement

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.

Explanation

Here's a breakdown of the solution:

• Frequency Analysis & Sorting: Calculate the frequency of each character in the input string s. Sort these frequencies in descending order. The highest frequencies will contribute most to the beauty of a k-subsequence.

• Maximize Beauty & Count Subsequences: To maximize the beauty, greedily select characters with the highest frequencies. Iterate through the sorted frequencies. Include characters with frequencies greater than a threshold until k characters are selected. If the k-th largest frequency occurs multiple times, calculate the number of ways to choose exactly k characters from all occurrences of that frequency in s.

• Combinations (nCr): The number of k-subsequences with maximum beauty hinges on calculating combinations efficiently, modulo 10⁹ + 7. Precompute factorials and their modular inverses for faster computation.

• Complexity Analysis:

  • Time Complexity: O(N + M log M), where N is the length of the string s and M is the number of unique characters (at most 26). Sorting the frequencies dominates.
  • Space Complexity: O(N + M), where N is the size of factorial cache and M is the number of unique characters, to store frequency counts and factorials.

Code

    def solve():
    s = input()
    k = int(input())

    MOD = 10**9 + 7

    counts = {}
    for char in s:
        counts[char] = counts.get(char, 0) + 1

    frequencies = sorted(counts.values(), reverse=True)

    max_beauty = sum(frequencies[:k])


    def combinations(n, r, factorial, inverse_factorial):
        if r < 0 or r > n:
            return 0
        if r == 0 or r == n:
            return 1
        if r > n // 2:
            r = n - r

        return (factorial[n] * inverse_factorial[r] % MOD) * inverse_factorial[n - r] % MOD


    max_freq = frequencies[k-1]
    count_max_freq = frequencies.count(max_freq)

    n = 0
    for char in counts:
        if counts[char] == max_freq:
            n += 1

    m = frequencies[:k].count(max_freq)

    N = len(s) + 1
    factorial = [1] * N
    for i in range(2, N):
        factorial[i] = (factorial[i - 1] * i) % MOD

    inverse_factorial = [1] * N
    inverse_factorial[N - 1] = pow(factorial[N - 1], MOD - 2, MOD)
    for i in range(N - 2, -1, -1):
        inverse_factorial[i] = (inverse_factorial[i + 1] * (i + 1)) % MOD


    ans = combinations(sum([1 for x in counts.values() if x == max_freq]), frequencies[:k].count(max_freq), factorial, inverse_factorial)
    print(ans % MOD)

solve()