2559. Count Vowel Strings in Ranges

Problem Description

The problem presents us with an array of strings words and a two-dimensional array queries. We need to determine how many strings within a specific index range in the words array start and end with a vowel. A string is said to start and end with a vowel if the first and the last character are both vowels. The range for each query is defined by two indices [l_i, r_i], where l_i is the start index and r_i is the end index. Our goal is to answer each query by counting the relevant strings and returning the results in an array ans, where each element ans[i] corresponds to the count for the i-th query.

The vowel characters are specifically 'a', 'e', 'i', 'o', and 'u'. The array words is 0-indexed, meaning the first element has an index of 0, the second an index of 1, and so on. This indexing is important to consider when processing the ranges specified in each query within queries.

Intuition

To efficiently solve this problem, we leverage two techniques:

1. Precomputation with Cumulative Sums: The idea is to precompute the number of valid strings (strings that start and end with a vowel) up to each index in the words array. This is achieved by creating an auxiliary array s where each element s[i] stores the cumulative count of valid strings from words[0] to words[i-1]. This allows us to quickly calculate the number of valid strings in any query range with a simple subtraction: s[r + 1] - s[l], where l and r are the start and end indices of the query.

2. Checking String Endpoints: For each string in words, we only need to check the first and the last character to determine if it qualifies as a valid string. We define the set of vowels and then, for each word, check if both endpoints are in that set. The Python expression int(w[0] in vowels and w[-1] in vowels) evaluates to 1 if both conditions are true and 0 otherwise.

By combining these two ideas, the solution efficiently computes the answer to each query by referencing the precomputed cumulative sum array, resulting in a significant reduction of the time complexity, especially when dealing with a large number of queries over the words array.

Learn more about Prefix Sum patterns.

Solution Approach

The Solution class has a method named vowelStrings which takes in two parameters: words, a list of strings, and queries, a 2D list of integers. The output is a list of integers representing the counts of strings in words that satisfy the query conditions (start and end with a vowel).

Here's a closer look at how the solution is implemented:

1. Defining a Set of Vowels: A set vowels is created with all the vowel characters {'a', 'e', 'i', 'o', 'u'}. This allows for constant time complexity O(1) for checking if a character is a vowel since set membership checking is typically O(1) due to hashing.

2. Calculating Cumulative Sums with accumulate: The accumulate function from Python's itertools module is used to generate the cumulative sum of valid strings. For every word w in words, an expression int(w[0] in vowels and w[-1] in vowels) is computed which results in 1 if the word starts and ends with a vowel and 0 otherwise. These values are accumulated to get cumulative sums. The initial=0 parameter is used to start the accumulation from 0 which implicitly adds an extra 0 at the beginning of the accumulated list, making the cumulative sum array s one element longer than the words. This simplification helps in handling edge cases and simplifies the subtraction step in the queries.

3. Handling Queries: With the cumulative sums array s ready, each query [l, r] is processed to find the count of valid strings in the range by subtracting the cumulative sum up to l from the cumulative sum up to r + 1. This operation takes constant time O(1) for each query. The result of this subtraction gives the count of valid strings within the specified range of indices, which is then appended to the resulting list.

4. Return the Results: A list comprehension is used to iterate through each query in queries, and the described subtraction operation is applied to produce the final answer list, which is then returned.

The overall time complexities for initializing the cumulative sum and for processing each query are O(N) and O(1) respectively, where N is the total number of words in words. Given Q queries, the total time complexity for the query processing part is O(Q). Hence the entire solution has an overall time complexity of O(N + Q) which is highly efficient, especially for large datasets.

All in all, this approach utilizes space-time trade-offs (precomputing the cumulative sum) and efficient data structure access patterns (using a set and direct indexing) to provide an optimal solution to the problem.

Example Walkthrough

Let's go through a walkthrough example to illustrate the solution approach. Suppose we are given an array of strings words and a 2D array queries as follows:

• words: ["apple", "orange", "coop", "peach", "echo"]
• queries: [[1, 3], [0, 4], [2, 2]]

First, we will start by identifying which words start and end with a vowel:

• "apple" starts with 'a' and ends with 'e', so it's valid.
• "orange" starts with 'o' and ends with 'e', so it's valid.
• "coop" does not end with a vowel, so it's not valid.
• "peach" does not start with a vowel, so it's not valid.
• "echo" starts with 'e' and ends with 'o', so it's valid.

Next, we create a cumulative sum array s which will hold the number of valid strings at each index. It will look like this:

• s: [0, 1, 2, 2, 2, 3]

Now, we process each query:

For the query [1, 3], we subtract the cumulative sum at index 1 from the cumulative sum at index 3 + 1 (since s has an extra zero at the beginning):

• s[3 + 1] - s[1] = s[4] - s[1] = 2 - 1 = 1. So there is one valid string in the range from index 1 to 3 in words.

For the query [0, 4], we do:

• s[4 + 1] - s[0] = s[5] - s[0] = 3 - 0 = 3. There are three valid strings in the range from index 0 to 4 in words.

For the query [2, 2], we do:

• s[2 + 1] - s[2] = s[3] - s[2] = 2 - 2 = 0. There are no valid strings at index 2 in words, because index 2 to 2 is the string "coop" which does not start and end with a vowel.

Therefore, the final result (counts of valid strings for each query) is:

• ans: [1, 3, 0]

We have used the vowels set to identify valid strings, the accumulate function to prepare the cumulative sum array s, and simple subtraction for each query to get the counts efficiently. This outlines the solution approach and demonstrates how it would work with an actual set of inputs.

Solution Implementation

Time and Space Complexity

Time Complexity

The time complexity of the code mainly comes from three parts:

1. Creating the list comprehension which checks every word in the input list words. This process is O(n) where n is the number of words in words.
2. The use of accumulate from the itertools module to create a prefix sum array of the boolean values obtained from the list comprehension. Since accumulate goes through each element in the provided list, this is O(n).
3. Finally, answering the queries involves iterating over each query and performing constant-time subtraction for the prefix sums. If there are q queries, this part has a complexity of O(q).

Combining these gives a total time complexity of O(n + q).

Space Complexity

The space complexity considerations include:

1. Storage of the boolean list comprehension which requires O(n) space.
2. The prefix sum list s which also needs O(n) space.
3. The space required for the output list which will store q integers, hence O(q) space.

The total space complexity is thus the sum of the above, which is O(n + q).

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