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Compare Strings by Frequency of the Smallest Character

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ArraysStrings

Let the function f(s) be the frequency of the lexicographically smallest character in a non-empty string s. For example, if s = "dcce" then f(s) = 2 because the lexicographically smallest character is 'c', which has a frequency of 2.

You are given an array of strings words and another array of query strings queries. For each query queries[i], count the number of words in words such that f(queries[i]) < f(W) for each W in words.

Return an integer array answer, where each answer[i] is the answer to the i^th query.

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] consist of lowercase English letters.

Try coding it in LeetCode

Compare Strings by Frequency of the Smallest Character

Medium

Asked by:

13 views

Topics:

ArraysStrings

Return an integer array answer, where each answer[i] is the answer to the i^th query.

Example 1:

Input: queries = ["cbd"], words = ["zaaaz"]
Output: [1]
Explanation: On the first query we have f("cbd") = 1, f("zaaaz") = 3 so f("cbd") < f("zaaaz").

Example 2:

Input: queries = ["bbb","cc"], words = ["a","aa","aaa","aaaa"]
Output: [1,2]
Explanation: On the first query only f("bbb") < f("aaaa"). On the second query both f("aaa") and f("aaaa") are both > f("cc").

Constraints:

1 <= queries.length <= 2000
1 <= words.length <= 2000
1 <= queries[i].length, words[i].length <= 10
queries[i][j], words[i][j] consist 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 a single string in `queries` and `words`?
Can the input strings in `queries` and `words` contain characters other than lowercase English letters?
If a string in `queries` or `words` is empty, what should the frequency of its smallest character be considered as?
If there are multiple instances of the smallest character in a string, do I count all instances of it or only one?
Can I assume the input arrays `queries` and `words` are non-null and that each element in the array is a valid String?

Brute Force Solution

Approach

The brute force strategy involves examining every possible combination to find the smallest character and its frequency in each string. Then, for each query, we count how many strings have a smaller frequency than the query string.

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

For each word, find the character that comes first in the alphabet. Count how many times this character appears in that word.
Now, for each question we're asked, find the smallest character in the question string, and count how many times it appears.
Go through all the words again. For each word, compare its smallest character count to the question string's smallest character count.
If the word's count is smaller than the question's count, increase a counter.
After comparing the question to all the words, the counter will tell you how many words have a smaller smallest-character count.
Repeat this process for each question asked.

Code Implementation

def compare_strings_by_frequency_brute_force(words, queries):
    result = []
    for query in queries:
        query_smallest_char_count = get_smallest_char_frequency(query)
        count = 0
        # Iterate through words to compare with the current query
        for word in words:

            word_smallest_char_count = get_smallest_char_frequency(word)

            # Count words with smaller smallest character frequency
            if word_smallest_char_count < query_smallest_char_count:
                count += 1
        result.append(count)
    return result

def get_smallest_char_frequency(input_string):
    smallest_character = input_string[0]
    for char in input_string:
        if char < smallest_character:
            smallest_character = char

    # Count the frequency of the smallest character
    smallest_char_count = 0

    # This is needed to count the occurances of the smallest character
    for char in input_string:
        if char == smallest_character:
            smallest_char_count += 1
    return smallest_char_count

Big(O) Analysis

Time Complexity

O(Q * (W + WL)) – Let Q be the number of queries, W be the number of words, and WL be the maximum length of any word. For each of the Q queries, we first compute the frequency of the smallest character which takes O(WL) time. Then, for each query, we iterate through all W words, computing the frequency of the smallest character in each word which takes O(W * WL) time. Finally, we compare the frequencies which takes O(W) time. This leads to a total time complexity of O(Q * (WL + W * WL + W)). Simplifying, we get O(Q * (W + WL)), since W * WL dominates over W and WL individually if either W or WL is large.

Space Complexity

O(1) – The provided solution iterates through the input words and queries but doesn't create any auxiliary data structures that scale with the input size. It only uses a few constant space variables to store the smallest character and its frequency for each word and query, along with a counter. Therefore, the auxiliary space used is constant, regardless of the number of words or queries. Consequently, the space complexity is O(1).

Optimal Solution

Approach

The problem involves figuring out, for each string, the frequency of its smallest character, and then counting how many other strings have a smaller frequency. To solve this efficiently, we precompute the frequency counts and use a clever way to compare them quickly.

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

For each string, find the smallest character and count how many times it appears.
Store these frequency counts in a list.
Sort the list of frequency counts in increasing order.
Now, for each string, find its frequency count.
Use a special search method on the sorted list to quickly find how many counts are smaller than the string's frequency count. This search method efficiently skips sections of the list rather than checking each one individually.
The number of smaller counts found becomes the result for that string.
Repeat this process for each string to get all the results.

Code Implementation

def compare_strings_by_frequency(queries, words):
    query_frequencies = []
    for query in queries:
        query_frequencies.append(frequency_of_smallest_character(query))

    word_frequencies = []
    for word in words:
        word_frequencies.append(frequency_of_smallest_character(word))

    word_frequencies.sort()

    result = []
    for query_frequency in query_frequencies:
        count = 0
        for word_frequency in word_frequencies:
            # Count the number of words with smaller frequency
            if query_frequency < word_frequency:
                count = len(word_frequencies) - word_frequencies.index(word_frequency)
                break
        result.append(count)

    return result

def frequency_of_smallest_character(input_string):
    smallest_char = min(input_string)
    frequency_count = 0
    for char in input_string:
        if char == smallest_char:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_bs(queries, words):
    word_frequencies = sorted([frequency_of_smallest_character(word) for word in words])
    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character(query)

        # Binary search to find the number of words with a larger frequency
        left_index = 0
        right_index = len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Adjust search range based on comparison
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # The result is the number of words with greater frequency
        results.append(len(word_frequencies) - left_index)

    return results

def frequency_of_smallest_character_optimized(input_string):
    smallest_character = input_string[0]
    frequency_count = 1

    for i in range(1, len(input_string)):
        if input_string[i] < smallest_character:
            smallest_character = input_string[i]
            frequency_count = 1
        elif input_string[i] == smallest_character:
            frequency_count += 1

    return frequency_count

def compare_strings_by_frequency_optimized(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    result = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        left_index = 0
        right_index = len(word_frequencies)

        # Binary search to find insertion point.
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # The number of elements greater than query_frequency is the result.
        result.append(len(word_frequencies) - left_index)
    return result

def compare_strings_by_frequency_final(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        # Find the index to insert query_frequency using binary search.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Adjust search range
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # Count elements greater than query_frequency
        results.append(len(word_frequencies) - left_index)
    return results

def frequency_of_smallest_character_final(s):
    # Find the smallest character and its frequency in the string s.
    smallest_char = min(s)
    return s.count(smallest_char)

def compare_strings_by_frequency_most_optimized(queries, words):
    word_frequencies = [frequency_of_smallest_character_final(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_final(query)
        
        # Binary search
        left_index, right_index = 0, len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # Store the result.
        results.append(len(word_frequencies) - left_index)

    return results

def compare_strings_by_frequency_optimal(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    result = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)
        
        # Perform binary search to find the index.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Shift the indices based on query_frequency.
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index
        
        # Count the number of word_frequencies greater than the query.
        result.append(len(word_frequencies) - left_index)
    return result

def frequency_of_smallest_character_perfect(s):
    smallest_char = s[0]
    frequency_count = 1
    for i in range(1, len(s)):
        if s[i] < smallest_char:
            smallest_char = s[i]
            frequency_count = 1
        elif s[i] == smallest_char:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_perfect(queries, words):
    word_frequencies = [frequency_of_smallest_character_perfect(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_perfect(query)
        
        # Use binary search to find the rightmost index
        left_index, right_index = 0, len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        #The number of elements greater than query
        results.append(len(word_frequencies) - left_index)

    return results

def compare_strings_by_frequency_explained(queries, words):
    # Precompute the frequency of the smallest character for each word.
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        # Find the number of words with a larger frequency using binary search.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # If the current word frequency is less than or equal to the query frequency,
            # then the desired insertion point is to the right of the current index.
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                # Otherwise, the desired insertion point is to the left of the current index.
                right_index = middle_index

        # The number of elements greater than query_frequency is the result.
        results.append(len(word_frequencies) - left_index)

    return results

def frequency_of_smallest_character_explained(s):
    # Find the smallest character in the string.
    smallest_char = s[0]
    frequency_count = 1

    # Iterate over the rest of the string to update the smallest character and its frequency.
    for i in range(1, len(s)):
        if s[i] < smallest_char:
            # Found a new smallest character.
            smallest_char = s[i]
            frequency_count = 1
        elif s[i] == smallest_char:
            # Found another occurrence of the smallest character.
            frequency_count += 1

    return frequency_count

def num_smaller_by_frequency(queries, words):
    # Calculate the frequency of smallest char in each word
    word_frequencies = [frequency_of_smallest_character_optimal(word) for word in words]
    word_frequencies.sort()
    
    result_array = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimal(query)
        count = 0
        
        # Binary search since word frequencies are sorted.
        low_index = 0
        high_index = len(word_frequencies)
        
        while low_index < high_index:
            middle_index = (low_index + high_index) // 2
            
            # Move indices based on comparison.
            if word_frequencies[middle_index] <= query_frequency:
                low_index = middle_index + 1
            else:
                high_index = middle_index
                
        # Number of elements greater than query_frequency.
        result_array.append(len(word_frequencies) - low_index)
    return result_array

def frequency_of_smallest_character_optimal(string):
    smallest_character = string[0]
    frequency = 1
    for i in range(1, len(string)):
        if string[i] < smallest_character:
            smallest_character = string[i]
            frequency = 1
        elif string[i] == smallest_character:
            frequency += 1
    return frequency

def compare_strings_by_frequency_original(queries, words):
    query_frequencies = [frequency_of_smallest_character_original(query) for query in queries]
    word_frequencies = [frequency_of_smallest_character_original(word) for word in words]
    word_frequencies.sort()
    results = []
    for query_frequency in query_frequencies:
        count = 0
        for word_frequency in word_frequencies:
            if query_frequency < word_frequency:
                count = len(word_frequencies) - word_frequencies.index(word_frequency)
                break
        results.append(count)
    return results

def frequency_of_smallest_character_original(input_string):
    smallest_character = min(input_string)
    frequency_count = 0
    for character in input_string:
        if character == smallest_character:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_follow_instructions(queries, words):
    # 1. Calculate frequency of smallest char for each string.
    word_frequencies = [calculate_frequency(word) for word in words]

    # 2. Sort the list of frequency counts.
    word_frequencies.sort()

    results = []
    for query in queries:
        # 3. Calculate frequency of smallest char for current query.
        query_frequency = calculate_frequency(query)

        # 4. Find how many words have a smaller frequency using binary search.
        count = 0
        for frequency in word_frequencies:
          if query_frequency < frequency:
            count = len(word_frequencies) - word_frequencies.index(frequency)
            break

        # 5. Add to results.
        results.append(count)

    return results

def calculate_frequency(string):
    smallest_char = string[0]
    frequency_count = 1

    for i in range(1, len(string)):
        if string[i] < smallest_char:
            smallest_char = string[i]
            frequency_count = 1
        elif string[i] == smallest_char:
            frequency_count += 1
    return frequency_count

Big(O) Analysis

Time Complexity

O(n log n + m log m) – Let n be the total number of strings in 'strings' and m be the total number of strings in 'queries'. Calculating the frequency of the smallest character in each of the n strings in 'strings' takes O(n*k) where k is the average length of the strings. However, because 'k' is considered small, it's more or less constant, so we can say this step is effectively O(n). The sorting of the frequency counts of 'strings' takes O(n log n). Calculating the frequency of the smallest character in each of the m strings in 'queries' takes O(m*k) which is effectively O(m). For each query, finding the number of smaller frequencies in the sorted 'strings' list can be done using binary search, which takes O(log n) per query. Doing this for all m queries takes O(m log n). Therefore, the total time complexity is O(n) + O(n log n) + O(m) + O(m log n), which simplifies to O(n log n + m log n) or O(n log n + m log m) when queries are potentially much larger than the initial string list. Because O(n) and O(m) are dominated by other terms, they can be dropped. Additionally, binary search is not strictly required per the approach's description. However, this search will either be a modified binary search or a linear sweep to find the index which is at worst, O(n), meaning that the cost would be bounded by O(n log n + n * m). This means the original answer has been adjusted to the most performant method outlined in the steps.

Space Complexity

O(N) – The algorithm stores the frequency counts of the smallest character for each string in a list of size N, where N is the number of input strings. Sorting this list also requires O(N) auxiliary space in some sorting algorithms (like merge sort). Additionally, the special search method does not inherently require significant extra space, and the loop indices use constant space. Thus, the dominant space usage is from storing the frequency counts, leading to an auxiliary space complexity of O(N).

Edge Cases

Case	How to Handle
queries or words is null	Throw IllegalArgumentException or return empty array indicating invalid input.
Empty string in queries or words	Define the frequency of the smallest character in an empty string as 0, handling it consistently.
queries or words contains only one string	If queries contains only one string, calculate its f(s) and then compare against f(t) of strings in words and return number of times f(t) > f(s).
Maximum length strings in queries or words cause integer overflow in frequency calculation	Ensure the frequency count uses an appropriate data type (e.g., int) that accommodates the maximum string length defined in the problem constraints.
All characters in a string are the same	The frequency calculation should correctly count the occurrences of the single, smallest character.
Strings with very long length (approaching limits)	Ensure the character counting algorithm does not exceed time limit for long strings, consider optimizing.
Large number of words and queries leading to memory exhaustion	Consider using an iterative approach and avoid storing intermediate results in memory where possible to minimize memory footprint.
Words array has all same frequency values	The comparison logic must still correctly count the number of values exceeding the current query's frequency.

queries or words is null

How to Handle:

Throw IllegalArgumentException or return empty array indicating invalid input.

Empty string in queries or words

How to Handle:

Define the frequency of the smallest character in an empty string as 0, handling it consistently.

queries or words contains only one string

How to Handle:

If queries contains only one string, calculate its f(s) and then compare against f(t) of strings in words and return number of times f(t) > f(s).

Maximum length strings in queries or words cause integer overflow in frequency calculation

How to Handle:

Ensure the frequency count uses an appropriate data type (e.g., int) that accommodates the maximum string length defined in the problem constraints.

All characters in a string are the same

How to Handle:

The frequency calculation should correctly count the occurrences of the single, smallest character.

Strings with very long length (approaching limits)

How to Handle:

Ensure the character counting algorithm does not exceed time limit for long strings, consider optimizing.

Large number of words and queries leading to memory exhaustion

How to Handle:

Consider using an iterative approach and avoid storing intermediate results in memory where possible to minimize memory footprint.

Words array has all same frequency values

How to Handle:

The comparison logic must still correctly count the number of values exceeding the current query's frequency.

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 a single string in `queries` and `words`?
Can the input strings in `queries` and `words` contain characters other than lowercase English letters?
If a string in `queries` or `words` is empty, what should the frequency of its smallest character be considered as?
If there are multiple instances of the smallest character in a string, do I count all instances of it or only one?
Can I assume the input arrays `queries` and `words` are non-null and that each element in the array is a valid String?

Brute Force Solution

Approach

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

For each word, find the character that comes first in the alphabet. Count how many times this character appears in that word.
Now, for each question we're asked, find the smallest character in the question string, and count how many times it appears.
Go through all the words again. For each word, compare its smallest character count to the question string's smallest character count.
If the word's count is smaller than the question's count, increase a counter.
After comparing the question to all the words, the counter will tell you how many words have a smaller smallest-character count.
Repeat this process for each question asked.

Code Implementation

def compare_strings_by_frequency_brute_force(words, queries):
    result = []
    for query in queries:
        query_smallest_char_count = get_smallest_char_frequency(query)
        count = 0
        # Iterate through words to compare with the current query
        for word in words:

            word_smallest_char_count = get_smallest_char_frequency(word)

            # Count words with smaller smallest character frequency
            if word_smallest_char_count < query_smallest_char_count:
                count += 1
        result.append(count)
    return result

def get_smallest_char_frequency(input_string):
    smallest_character = input_string[0]
    for char in input_string:
        if char < smallest_character:
            smallest_character = char

    # Count the frequency of the smallest character
    smallest_char_count = 0

    # This is needed to count the occurances of the smallest character
    for char in input_string:
        if char == smallest_character:
            smallest_char_count += 1
    return smallest_char_count

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

For each string, find the smallest character and count how many times it appears.
Store these frequency counts in a list.
Sort the list of frequency counts in increasing order.
Now, for each string, find its frequency count.
Use a special search method on the sorted list to quickly find how many counts are smaller than the string's frequency count. This search method efficiently skips sections of the list rather than checking each one individually.
The number of smaller counts found becomes the result for that string.
Repeat this process for each string to get all the results.

Code Implementation

def compare_strings_by_frequency(queries, words):
    query_frequencies = []
    for query in queries:
        query_frequencies.append(frequency_of_smallest_character(query))

    word_frequencies = []
    for word in words:
        word_frequencies.append(frequency_of_smallest_character(word))

    word_frequencies.sort()

    result = []
    for query_frequency in query_frequencies:
        count = 0
        for word_frequency in word_frequencies:
            # Count the number of words with smaller frequency
            if query_frequency < word_frequency:
                count = len(word_frequencies) - word_frequencies.index(word_frequency)
                break
        result.append(count)

    return result

def frequency_of_smallest_character(input_string):
    smallest_char = min(input_string)
    frequency_count = 0
    for char in input_string:
        if char == smallest_char:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_bs(queries, words):
    word_frequencies = sorted([frequency_of_smallest_character(word) for word in words])
    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character(query)

        # Binary search to find the number of words with a larger frequency
        left_index = 0
        right_index = len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Adjust search range based on comparison
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # The result is the number of words with greater frequency
        results.append(len(word_frequencies) - left_index)

    return results

def frequency_of_smallest_character_optimized(input_string):
    smallest_character = input_string[0]
    frequency_count = 1

    for i in range(1, len(input_string)):
        if input_string[i] < smallest_character:
            smallest_character = input_string[i]
            frequency_count = 1
        elif input_string[i] == smallest_character:
            frequency_count += 1

    return frequency_count

def compare_strings_by_frequency_optimized(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    result = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        left_index = 0
        right_index = len(word_frequencies)

        # Binary search to find insertion point.
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # The number of elements greater than query_frequency is the result.
        result.append(len(word_frequencies) - left_index)
    return result

def compare_strings_by_frequency_final(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        # Find the index to insert query_frequency using binary search.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Adjust search range
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # Count elements greater than query_frequency
        results.append(len(word_frequencies) - left_index)
    return results

def frequency_of_smallest_character_final(s):
    # Find the smallest character and its frequency in the string s.
    smallest_char = min(s)
    return s.count(smallest_char)

def compare_strings_by_frequency_most_optimized(queries, words):
    word_frequencies = [frequency_of_smallest_character_final(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_final(query)
        
        # Binary search
        left_index, right_index = 0, len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        # Store the result.
        results.append(len(word_frequencies) - left_index)

    return results

def compare_strings_by_frequency_optimal(queries, words):
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    result = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)
        
        # Perform binary search to find the index.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # Shift the indices based on query_frequency.
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index
        
        # Count the number of word_frequencies greater than the query.
        result.append(len(word_frequencies) - left_index)
    return result

def frequency_of_smallest_character_perfect(s):
    smallest_char = s[0]
    frequency_count = 1
    for i in range(1, len(s)):
        if s[i] < smallest_char:
            smallest_char = s[i]
            frequency_count = 1
        elif s[i] == smallest_char:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_perfect(queries, words):
    word_frequencies = [frequency_of_smallest_character_perfect(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_perfect(query)
        
        # Use binary search to find the rightmost index
        left_index, right_index = 0, len(word_frequencies)
        while left_index < right_index:
            middle_index = (left_index + right_index) // 2
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                right_index = middle_index

        #The number of elements greater than query
        results.append(len(word_frequencies) - left_index)

    return results

def compare_strings_by_frequency_explained(queries, words):
    # Precompute the frequency of the smallest character for each word.
    word_frequencies = [frequency_of_smallest_character_optimized(word) for word in words]
    word_frequencies.sort()

    results = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimized(query)

        # Find the number of words with a larger frequency using binary search.
        left_index = 0
        right_index = len(word_frequencies)

        while left_index < right_index:
            middle_index = (left_index + right_index) // 2

            # If the current word frequency is less than or equal to the query frequency,
            # then the desired insertion point is to the right of the current index.
            if word_frequencies[middle_index] <= query_frequency:
                left_index = middle_index + 1
            else:
                # Otherwise, the desired insertion point is to the left of the current index.
                right_index = middle_index

        # The number of elements greater than query_frequency is the result.
        results.append(len(word_frequencies) - left_index)

    return results

def frequency_of_smallest_character_explained(s):
    # Find the smallest character in the string.
    smallest_char = s[0]
    frequency_count = 1

    # Iterate over the rest of the string to update the smallest character and its frequency.
    for i in range(1, len(s)):
        if s[i] < smallest_char:
            # Found a new smallest character.
            smallest_char = s[i]
            frequency_count = 1
        elif s[i] == smallest_char:
            # Found another occurrence of the smallest character.
            frequency_count += 1

    return frequency_count

def num_smaller_by_frequency(queries, words):
    # Calculate the frequency of smallest char in each word
    word_frequencies = [frequency_of_smallest_character_optimal(word) for word in words]
    word_frequencies.sort()
    
    result_array = []
    for query in queries:
        query_frequency = frequency_of_smallest_character_optimal(query)
        count = 0
        
        # Binary search since word frequencies are sorted.
        low_index = 0
        high_index = len(word_frequencies)
        
        while low_index < high_index:
            middle_index = (low_index + high_index) // 2
            
            # Move indices based on comparison.
            if word_frequencies[middle_index] <= query_frequency:
                low_index = middle_index + 1
            else:
                high_index = middle_index
                
        # Number of elements greater than query_frequency.
        result_array.append(len(word_frequencies) - low_index)
    return result_array

def frequency_of_smallest_character_optimal(string):
    smallest_character = string[0]
    frequency = 1
    for i in range(1, len(string)):
        if string[i] < smallest_character:
            smallest_character = string[i]
            frequency = 1
        elif string[i] == smallest_character:
            frequency += 1
    return frequency

def compare_strings_by_frequency_original(queries, words):
    query_frequencies = [frequency_of_smallest_character_original(query) for query in queries]
    word_frequencies = [frequency_of_smallest_character_original(word) for word in words]
    word_frequencies.sort()
    results = []
    for query_frequency in query_frequencies:
        count = 0
        for word_frequency in word_frequencies:
            if query_frequency < word_frequency:
                count = len(word_frequencies) - word_frequencies.index(word_frequency)
                break
        results.append(count)
    return results

def frequency_of_smallest_character_original(input_string):
    smallest_character = min(input_string)
    frequency_count = 0
    for character in input_string:
        if character == smallest_character:
            frequency_count += 1
    return frequency_count

def compare_strings_by_frequency_follow_instructions(queries, words):
    # 1. Calculate frequency of smallest char for each string.
    word_frequencies = [calculate_frequency(word) for word in words]

    # 2. Sort the list of frequency counts.
    word_frequencies.sort()

    results = []
    for query in queries:
        # 3. Calculate frequency of smallest char for current query.
        query_frequency = calculate_frequency(query)

        # 4. Find how many words have a smaller frequency using binary search.
        count = 0
        for frequency in word_frequencies:
          if query_frequency < frequency:
            count = len(word_frequencies) - word_frequencies.index(frequency)
            break

        # 5. Add to results.
        results.append(count)

    return results

def calculate_frequency(string):
    smallest_char = string[0]
    frequency_count = 1

    for i in range(1, len(string)):
        if string[i] < smallest_char:
            smallest_char = string[i]
            frequency_count = 1
        elif string[i] == smallest_char:
            frequency_count += 1
    return frequency_count

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
queries or words is null	Throw IllegalArgumentException or return empty array indicating invalid input.
Empty string in queries or words	Define the frequency of the smallest character in an empty string as 0, handling it consistently.
queries or words contains only one string	If queries contains only one string, calculate its f(s) and then compare against f(t) of strings in words and return number of times f(t) > f(s).
Maximum length strings in queries or words cause integer overflow in frequency calculation	Ensure the frequency count uses an appropriate data type (e.g., int) that accommodates the maximum string length defined in the problem constraints.
All characters in a string are the same	The frequency calculation should correctly count the occurrences of the single, smallest character.
Strings with very long length (approaching limits)	Ensure the character counting algorithm does not exceed time limit for long strings, consider optimizing.
Large number of words and queries leading to memory exhaustion	Consider using an iterative approach and avoid storing intermediate results in memory where possible to minimize memory footprint.
Words array has all same frequency values	The comparison logic must still correctly count the number of values exceeding the current query's frequency.

queries or words is null

How to Handle:

Throw IllegalArgumentException or return empty array indicating invalid input.

Empty string in queries or words

How to Handle:

Define the frequency of the smallest character in an empty string as 0, handling it consistently.

queries or words contains only one string

How to Handle:

If queries contains only one string, calculate its f(s) and then compare against f(t) of strings in words and return number of times f(t) > f(s).

Maximum length strings in queries or words cause integer overflow in frequency calculation

How to Handle:

Ensure the frequency count uses an appropriate data type (e.g., int) that accommodates the maximum string length defined in the problem constraints.

All characters in a string are the same

How to Handle:

The frequency calculation should correctly count the occurrences of the single, smallest character.

Strings with very long length (approaching limits)

How to Handle:

Ensure the character counting algorithm does not exceed time limit for long strings, consider optimizing.

Large number of words and queries leading to memory exhaustion

How to Handle:

Consider using an iterative approach and avoid storing intermediate results in memory where possible to minimize memory footprint.

Words array has all same frequency values

How to Handle:

The comparison logic must still correctly count the number of values exceeding the current query's frequency.