Problem Description

You need to understand a special function f(s) that counts how many times the alphabetically smallest character appears in a string s. For example, in the string "dcce", the smallest character is 'c' and it appears 2 times, so f("dcce") = 2.

Given two arrays:

• words: an array of strings
• queries: an array of query strings

For each query string in queries, you need to count how many strings in words have a higher f value than the query string's f value.

In other words, for each queries[i], count how many words W in words satisfy the condition: f(queries[i]) < f(W).

Return an array answer where answer[i] contains the count for the i-th query.

Example breakdown:

• If queries[0] = "abc" with f("abc") = 1 (character 'a' appears once)
• And words = ["aa", "aaa", "aaaa"] with f values [2, 3, 4] respectively
• All three words have f values greater than 1, so answer[0] = 3

The solution uses sorting and binary search to efficiently find how many words have f values greater than each query's f value.

Intuition

The key observation is that we need to repeatedly answer "how many values are greater than X?" for different X values. This is a classic pattern that suggests preprocessing and binary search.

If we process each query naively by comparing it against all words every time, we'd have O(queries.length * words.length) comparisons, which could be inefficient for large inputs.

Instead, we can precompute all f values for the words array once and sort them. Why sorting? Because once sorted, all values greater than a certain threshold will be grouped together at the end of the array. This allows us to use binary search to quickly find where this group starts.

For example, if we have sorted f values [1, 2, 2, 3, 5, 7] and we want to count how many are greater than 2, we can use binary search to find that values greater than 2 start at index 3. Therefore, there are 6 - 3 = 3 values greater than 2.

The function bisect_right finds the rightmost position where we can insert a value while maintaining sorted order. This gives us the index where values greater than our query start. Subtracting this index from the total length gives us the count of greater values.

This preprocessing approach reduces our per-query time from O(words.length) to O(log(words.length)), making the solution much more efficient when handling multiple queries.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The solution implements the approach using sorting and binary search as outlined:

Step 1: Implement the function f(s)

The function uses Python's Counter to count character frequencies and iterates through lowercase letters in alphabetical order to find the first character that appears in the string:

def f(s: str) -> int:
    cnt = Counter(s)
    return next(cnt[c] for c in ascii_lowercase if cnt[c])

This guarantees we get the frequency of the lexicographically smallest character.

Step 2: Preprocess the words array

Calculate f(w) for each word w in words and sort these values:

n = len(words)
nums = sorted(f(w) for w in words)

After sorting, nums contains all f values in ascending order, making it suitable for binary search.

Step 3: Process each query

For each query q, we:

1. Calculate f(q)
2. Use bisect_right(nums, f(q)) to find the insertion point for f(q) in the sorted array
3. This insertion point tells us how many values are less than or equal to f(q)
4. Subtract from n to get the count of values greater than f(q)

return [n - bisect_right(nums, f(q)) for q in queries]

Time Complexity Analysis:

• Computing f for all words: O(m * k) where m is the number of words and k is the average word length
• Sorting: O(m * log(m))
• For each query: O(k + log(m)) where k is for computing f(q) and log(m) is for binary search
• Total: O(m * k + m * log(m) + p * (k + log(m))) where p is the number of queries

Space Complexity: O(m) for storing the sorted nums array.

Example Walkthrough

Let's walk through a concrete example to illustrate the solution approach.

Given:

• words = ["bbb", "cc", "a", "aa"]
• queries = ["aaa", "bb"]

Step 1: Calculate f values for words

For each word, we find the lexicographically smallest character and count its occurrences:

• f("bbb"): smallest char is 'b', appears 3 times → f = 3
• f("cc"): smallest char is 'c', appears 2 times → f = 2
• f("a"): smallest char is 'a', appears 1 time → f = 1
• f("aa"): smallest char is 'a', appears 2 times → f = 2

So our f values for words are: [3, 2, 1, 2]

Step 2: Sort the f values

After sorting: nums = [1, 2, 2, 3]

This sorted array enables efficient binary search to find how many values exceed any given threshold.

Step 3: Process each query

Query 1: "aaa"

• Calculate f("aaa"): smallest char is 'a', appears 3 times → f = 3
• Use binary search (bisect_right) to find where 3 would be inserted in [1, 2, 2, 3]
• bisect_right([1, 2, 2, 3], 3) = 4 (would insert after the existing 3)
• Count of values greater than 3: 4 - 4 = 0
• No words have f value greater than 3

Query 2: "bb"

• Calculate f("bb"): smallest char is 'b', appears 2 times → f = 2
• Use binary search to find where 2 would be inserted in [1, 2, 2, 3]
• bisect_right([1, 2, 2, 3], 2) = 3 (would insert after both 2's)
• Count of values greater than 2: 4 - 3 = 1
• One word ("bbb" with f=3) has f value greater than 2

Final Answer: [0, 1]

The key insight is that bisect_right tells us exactly how many elements are less than or equal to our target value. By subtracting this from the total count, we get the number of elements strictly greater than our target - which is exactly what we need for each query.

Solution Implementation

Time and Space Complexity

Time Complexity: O((n + q) × M + n × log n)

The time complexity consists of several parts:

• Computing f(w) for all words in words: Each call to f() takes O(M) time where M is the maximum string length (creating Counter and iterating through it). This is done n times, resulting in O(n × M).
• Sorting the nums array: O(n × log n) where n is the length of words.
• Computing f(q) for all queries and performing binary search: For each of the q queries, we compute f(q) in O(M) time and perform bisect_right in O(log n) time. This gives us O(q × M + q × log n).
• Total: O(n × M + n × log n + q × M + q × log n) = O((n + q) × M + n × log n + q × log n)

Since typically M < log n for reasonable inputs, the dominant term is O((n + q) × M + n × log n). However, the reference answer simplifies this to O((n + q) × M), which suggests either the sorting cost is considered negligible compared to string processing, or there's an assumption that n × log n is bounded by n × M.

Space Complexity: O(n)

The space complexity includes:

• nums array storing n frequency values: O(n)
• Temporary Counter objects created in function f(): O(26) = O(1) (only lowercase letters)
• Result list: O(q) (not counted in auxiliary space as it's the output)
• Total auxiliary space: O(n)

Learn more about how to find time and space complexity quickly.

Common Pitfalls

1. Incorrectly Finding the Lexicographically Smallest Character

A common mistake is using min(s) to find the smallest character, which works but then counting its frequency separately can be inefficient or error-prone:

Incorrect approach:

def f(s: str) -> int:
    smallest = min(s)
    return s.count(smallest)

While this works, iterating through the string multiple times (once for min, once for count) is less efficient than necessary.

Better approach:

def f(s: str) -> int:
    cnt = Counter(s)
    for c in ascii_lowercase:
        if cnt[c] > 0:
            return cnt[c]
    return 0

2. Using bisect_left Instead of bisect_right

This is a critical error that leads to incorrect counts. The difference matters when the query's f-value equals some word's f-value:

Incorrect:

position = bisect_left(word_frequencies, query_frequency)
return word_count - position

Why it's wrong: bisect_left finds the leftmost position to insert, so if query_frequency = 2 and word_frequencies = [1, 2, 2, 3], bisect_left returns 1, giving count = 4 - 1 = 3. But we want values strictly greater than 2, which should be 1 (only the value 3).

Correct:

position = bisect_right(word_frequencies, query_frequency)
return word_count - position

With bisect_right, we get position 3, giving count = 4 - 3 = 1, which is correct.

3. Not Sorting the Word Frequencies

Forgetting to sort the frequencies before using binary search will produce incorrect results:

Incorrect:

word_frequencies = [f(word) for word in words]  # Missing sort!
# Binary search on unsorted array gives wrong results

Correct:

word_frequencies = sorted(f(word) for word in words)

4. Recalculating f-values Multiple Times

Computing the f-value for words inside the query loop is inefficient:

Inefficient:

result = []
for query in queries:
    count = 0
    for word in words:
        if f(query) < f(word):  # Recalculating f(word) every time
            count += 1
    result.append(count)

Efficient: Pre-compute and store all f-values once, then reuse them for all queries.