Problem Description

The problem presents a function f(s) that calculates the frequency of the lexicographically smallest character within a non-empty string s. For instance, if s = "dcce", then f(s) equals 2 because the smallest character ('c') appears twice in the string.

Given two arrays, one containing strings words and the other containing query strings queries, the task is to determine how many words in words have a higher f value than the f value of each query string in queries. Specifically, for each query queries[i], you need to count the number of strings in words where f(queries[i]) is less than f(W) for each string W in words.

The expected output is an array answer, where answer[i] corresponds to the count for queries[i]. Essentially, the output tells us for each query, the count of words in words with a smaller lexicographically smallest character frequency.

Intuition

The intuition behind the solution involves first understanding what the function f does in terms of string processing. It counts the number of times the smallest lexicographical character appears in a string.

Approaching the solution, we consider these steps:

1. Precompute the f values for all strings in words since we will need to compare these with each query's f value. This precomputation helps to avoid recalculating these values for each query, which would be inefficient.

2. Sort the precomputed f values of the words array. With a sorted list of f values, we can efficiently determine where a particular value would be placed: all elements larger than this value would come after it in the sorted array.

3. For each query, calculate its f value. Then, to find out how many f values in words are larger than that of the query, we use binary search to find the insertion point in the sorted list of f values of words. This gives us the position where the f value of the query would fit if it were to be inserted into the list, effectively giving us the count of f values that are greater because all values to the right of this point are larger.

4. The bisect_right function from Python's bisect module is used to perform the binary search. It returns the insertion point (index) in the sorted nums list, which allows us to subtract this index from the total number of words to get the count of words with a larger f value than the query.

5. The final result for each query is calculated by taking the total number of words n and subtracting the insertion index. This count is added to the result list, which is returned at the end.

The process leverages binary search for efficiency, ensuring that an overall time complexity better than O(n^2) is achieved for the solution, where n is the length of the longest array between queries and words.

Learn more about Binary Search and Sorting patterns.

Solution Approach

The implementation of the solution to this problem uses Python's Counter class from the collections module, the ascii_lowercase string from the string module, and the bisect_right function from the bisect module. Here's how these are used step-by-step:

1. Define the function f(s): The function f(s) is where the calculation of the smallest character's frequency is implemented. It uses the Counter class to count the occurrences of each character in string s. Counter(s) returns a dictionary-like object where characters are keys, and their counts are values.

2. Find the smallest character's frequency: Once we have the counts, we iterate over ascii_lowercase (a string containing all lowercase letters in alphabetical order) to find the smallest character present in s. As soon as we find a character c that is in s (i.e., cnt[c] is not zero), we return its count cnt[c]. Using ascii_lowercase ensures we check characters in lexicographical order.

3. Precompute f for words: We create a list called nums that holds the f value for each word in words. This is done with a list comprehension [f(w) for w in words], calculating f(w) for each w in words.

4. Sort nums: nums is sorted so that binary search can be used to efficiently find positions within it. We use Python’s built-in sorted method for this purpose.

5. Find each query's count: For each query q in queries, we calculate f(q) and then find the index at which f(q) would be inserted while maintaining the order in nums. This is where bisect_right comes in. It returns the insertion point to the right of existing values of f(q). This index tells us how many words in words have a larger f value than f(q).

6. Prepare the result: Knowing the index from bisect_right, the number of words with a larger f value is found by subtracting this index from the total number of words, which is len(words). The comprehension [n - bisect_right(nums, f(q)) for q in queries] creates the result list by performing this calculation for each query.

7. Return the result: The resultant list is returned, which contains the count for each query indicating how many words in words have a higher f value than that of the query.

In conclusion, the solution makes use of a combination of string processing, precomputation, binary search, and efficient sorting to arrive at the result. It leverages the properties of sorted arrays and binary search (bisect_right) to perform queries in logarithmic time, after an initial sort cost, allowing for an efficient and scalable approach to solve the given problem.

Example Walkthrough

Let's go through an example to illustrate the solution approach described above. Suppose we have a list of words, words = ["a", "aa", "aaa", "aaaa"] and a list of queries, queries = ["a", "aa", "aaa"]. Our goal is to count how many words have a higher frequency of the lexicographically smallest character for each query. Let's walk through each step of the approach.

1. Define and calculate f for each word: We first define the function f(s) to calculate the frequency of the lexicographically smallest character in a given string s. For the words list, since all words consist of the character 'a', the f values are simply the lengths of the strings. Thus, f("a") = 1, f("aa") = 2, and so on.

2. Precompute f values for words: We precompute the f values for the words. This gives us nums = [1, 2, 3, 4].

3. Sort the nums list: We sort the precomputed f values of words. Our nums list is already sorted: nums = [1, 2, 3, 4].

4. Find each query's count: Now, we need to find out for each query how many f values in words are larger than the query's f value. For queries[i] = "a", f("a") is 1. Using binary search, we determine the position where 1 would fit in the sorted nums list (which is index 1, right after the first occurrence of 1). Since there are a total of 4 words, and 1 would be inserted at index 1, there are 4 - 1 = 3 words with a larger f value.

   Repeating this for queries[1] = "aa" with f("aa") = 2, binary search tells us that 2 would be inserted at index 2 in nums, leaving 4 - 2 = 2 words with a larger f value.

   Lastly, for queries[2] = "aaa", f("aaa") = 3, binary search gives us an insertion index of 3, so there are 4 - 3 = 1 word with a larger f value.

5. Prepare the result: The counts we determined in the previous step give us our final result array: [3, 2, 1].

6. Return the result: The result for our query is [3, 2, 1], indicating for each query in queries, how many words in words have a higher f value.

The example confirms our approach, showing how we can effectively calculate the counts required using precomputation, sorting, and binary search.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the provided code involves several components:

1. Calculating the frequency of the smallest character in each word using the f function. This operation involves creating a Counter for each word, which has a time complexity of O(m), where m is the average string length. This is done for all words, so for n words, this part would have a time complexity of O(n * m).

2. Sorting the frequencies of words. The sorted function has a time complexity of O(n log n), where n is the length of the words list.

3. For each query in queries, the function calculates the frequency of the smallest character and uses binary search (bisect_right) to find the position in the sorted list of word frequencies. The frequency calculation is O(m), and binary search in a sorted list of size n is O(log n). Therefore, for q queries, this part has a time complexity of O(q * (m + log n)).

Putting it all together, the total time complexity is O(n * m) + O(n log n) + O(q * (m + log n)).

Space Complexity

The space complexity is the additional space used by the algorithm:

1. Storing the frequencies of all words, which is O(n) space.

2. The sorted frequency list, which again takes O(n) space.

3. The counter created for each word, which in the worst case takes O(m) space; however, since the counters are not stored and are used one at a time, this is not multiplied by n.

4. The final result list which will contain q elements, resulting in O(q) space.

So, the total space complexity is O(n + q), since n and q space requirements dominate the O(m) space needed for the counter for each word or query string.

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