Count K-Subsequences of a String With Maximum Beauty

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StringsDynamic ProgrammingGreedy AlgorithmsArrays

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.

Try coding it in LeetCode

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the maximum length of the input string, and what are the possible characters that can appear in the string?
What are the constraints on the value of 'K', specifically, what is the maximum possible value of K?
If multiple K-subsequences have the same maximum beauty, should I return the count of all such subsequences, or is it sufficient to return just one?
How is the 'beauty' of a subsequence defined? (e.g., is it based on the frequency of characters, their positions, or some other criteria?)
If no K-subsequence exists, what should the function return (e.g., 0, -1, null)?

Brute Force Solution

Approach

We are given a string and need to find a certain number of subsequences that have the maximum possible beauty. The brute force method involves checking every single possible subsequence and calculating its beauty to see if it's among the most beautiful.

Here's how the algorithm would work step-by-step:

Consider every possible combination of characters from the string. This means looking at every single subsequence, even those that are very short or very long.
For each subsequence, calculate its beauty based on the frequency of each character in the subsequence.
Keep track of the beauty of each subsequence and which characters form that subsequence.
Find the highest beauty value among all the subsequences.
Count how many subsequences have this highest beauty value. These are the K-subsequences with maximum beauty.
If the number of subsequences with maximum beauty is greater than K, return K. Otherwise, return the actual count of subsequences with maximum beauty.

Code Implementation

def count_k_subsequences_brute_force(input_string, k_value):
    string_length = len(input_string)
    all_subsequences = []

    for i in range(1 << string_length):
        subsequence = ""
        for j in range(string_length):
            if (i >> j) & 1:
                subsequence += input_string[j]
        all_subsequences.append(subsequence)

    subsequence_beauties = []
    for subsequence in all_subsequences:
        character_frequencies = {}
        for char in subsequence:
            character_frequencies[char] = character_frequencies.get(char, 0) + 1

        subsequence_beauty = 0
        for char in character_frequencies:
            subsequence_beauty += character_frequencies[char] * character_frequencies[char]
        subsequence_beauties.append(subsequence_beauty)

    max_beauty = 0
    if subsequence_beauties:
        max_beauty = max(subsequence_beauties)

    max_beauty_count = 0

    # Must iterate through all subsequences and count
    for beauty in subsequence_beauties:
        if beauty == max_beauty:
            max_beauty_count += 1

    # Ensure to only return k
    if max_beauty_count > k_value:
        return k_value
    else:
        return max_beauty_count

Big(O) Analysis

Time Complexity

O(2^n) – The provided approach involves considering every possible subsequence of the input string of length n. Generating all subsequences requires examining each character and deciding whether to include it or exclude it in the current subsequence. This creates a binary choice for each of the n characters, resulting in 2^n possible subsequences. Calculating the beauty for each subsequence takes O(n) time in the worst case, but since the dominant factor is the generation of subsequences, the overall time complexity is O(2^n), as subsequence generation will be performed the same way regardless of string content or character counts.

Space Complexity

O(1) – The brute force approach, as described, doesn't explicitly use any auxiliary data structures that scale with the input string's length (N). It involves calculating beauty values and keeping track of the maximum beauty and its count; these can be done using a few constant-sized variables to store the maximum beauty, the count of subsequences with maximum beauty, and potentially some temporary variables for beauty calculation of individual subsequences. Therefore, the space complexity is constant. No lists, hash maps, or recursion that depend on the input size are present.

Optimal Solution

Approach

The best way to solve this problem is to figure out which letters give you the most 'beauty' and then take as many of those as possible. We do this by counting up the letters and prioritizing the most frequent ones to meet our subsequence limit.

Here's how the algorithm would work step-by-step:

First, count how many times each letter appears in the string.
Sort the letters based on how many times they appear, from most frequent to least frequent.
Then, pick the letters that appear most often, adding their counts together, until you either use all the letters or you reach the limit of how many letters we can pick.
If you still have letters available within the number allowed, add letters of decreasing frequency.
Keep track of how many times each of those most frequent letters appear.
To calculate the total 'beauty', multiply how many times each letter appears by itself and then add all of those results together.
The final sum is the maximum possible 'beauty' you can achieve.

Code Implementation

def count_k_subsequences(input_string, max_subsequence_length):
    character_counts = {}
    for char in input_string:
        character_counts[char] = character_counts.get(char, 0) + 1

    # Sort characters by frequency in descending order.
    sorted_characters = sorted(character_counts.items(), key=lambda item: item[1], reverse=True)

    total_beauty = 0
    total_characters_used = 0
    used_character_counts = {}

    # Greedily pick characters with the highest counts.
    for char, count in sorted_characters:
        if total_characters_used + count <= max_subsequence_length:
            used_character_counts[char] = count
            total_characters_used += count
        else:
            # Use only as many of this character as we can.
            remaining_space = max_subsequence_length - total_characters_used
            if remaining_space > 0:
                used_character_counts[char] = remaining_space
                total_characters_used = max_subsequence_length
            break

    # Calculate the beauty of the chosen subsequence.
    for char, count in used_character_counts.items():
        total_beauty += count * count

    return total_beauty

Big(O) Analysis

Time Complexity

O(n log n) – The first step involves counting the frequency of each character, which takes O(n) time where n is the length of the string. Next, we sort the character counts, which takes O(k log k) time, where k is the number of unique characters (at most 26 for lowercase English letters, and can be considered constant in many practical scenarios but we'll keep k for clarity). The subsequent steps of selecting letters and calculating beauty involve iterating through the sorted counts, taking O(k) time. Therefore, the dominant operation is the sorting, making the overall time complexity O(n + k log k). Since k is bounded by the alphabet size (a constant) and could potentially be proportional to n, we can say the worst-case complexity is O(n log n) since sorting n numbers will take n log n.

Space Complexity

O(1) – The space complexity is dominated by the frequency counting step. While a dictionary or array is implicitly used to count letter frequencies, the size is bounded by the number of possible characters, which is constant (26 for lowercase English letters). Sorting the frequencies is done in place or uses space proportional to the number of distinct characters which is a constant. Therefore, the auxiliary space used is constant regardless of the input string length N.

Edge Cases

Case	How to Handle
Empty string input	Return 0 immediately since no subsequences can be formed.
K is zero or negative	If K is zero, return 1 (empty subsequence). If K is negative, return 0 (no valid subsequences).
String length is less than K	Return 0, because there are not enough unique characters to form a k-length subsequence.
String contains characters with values that could cause integer overflow when multiplied.	Use modulo operator with a large prime number during multiplication to prevent overflow, applying the modulo to intermediate results as well.
K is larger than the number of unique characters in the string	Compute frequency of each character and consider only unique ones, adjusting the count if K exceeds this number.
All characters in the string are the same	If K > 1, return 0 (no solution). If K == 1, return 1 (only 1 subsequence). Handle this by checking if the character count equals the length of the input string.
String with very long length causing potential performance bottlenecks	Optimize character counting using efficient data structures (hash maps) and limit calculations to only relevant characters for k-subsequence creation.
String with extreme character frequencies causing integer overflow in subsequence count	Employ modulo operation at each step of the calculation, especially during multiplication of frequencies to prevent integer overflows.

Empty string input

How to Handle:

Return 0 immediately since no subsequences can be formed.

K is zero or negative

How to Handle:

If K is zero, return 1 (empty subsequence). If K is negative, return 0 (no valid subsequences).

String length is less than K

How to Handle:

Return 0, because there are not enough unique characters to form a k-length subsequence.

String contains characters with values that could cause integer overflow when multiplied.

How to Handle:

Use modulo operator with a large prime number during multiplication to prevent overflow, applying the modulo to intermediate results as well.

K is larger than the number of unique characters in the string

How to Handle:

Compute frequency of each character and consider only unique ones, adjusting the count if K exceeds this number.

All characters in the string are the same

How to Handle:

If K > 1, return 0 (no solution). If K == 1, return 1 (only 1 subsequence). Handle this by checking if the character count equals the length of the input string.

String with very long length causing potential performance bottlenecks

How to Handle:

Optimize character counting using efficient data structures (hash maps) and limit calculations to only relevant characters for k-subsequence creation.

String with extreme character frequencies causing integer overflow in subsequence count

How to Handle:

Employ modulo operation at each step of the calculation, especially during multiplication of frequencies to prevent integer overflows.