2559. Count Vowel Strings in Ranges - Detailed Explanation
Problem Statement
Description:
You are given an array of strings, words, and a list of queries, where each query is a pair of integers [l, r]. For each query, you need to count how many strings in the subarray words[l] to words[r] (inclusive) are “vowel strings.” A vowel string is defined as a string that both starts and ends with a vowel. The vowels are: a, e, i, o, and u.
Example 1:
- Input:
words = ["aba", "bcb", "ece", "aa", "e"]queries = [[0,2],[1,4],[1,1]]
- Output:
[2, 3, 0] - Explanation:
- For query
[0,2], the subarray is["aba", "bcb", "ece"]. Among these,"aba"and"ece"are vowel strings, so the count is 2. - For query
[1,4], the subarray is["bcb", "ece", "aa", "e"]. Here,"ece","aa", and"e"are vowel strings, so the count is 3. - For query
[1,1], the subarray is["bcb"], which does not start and end with a vowel, so the count is 0.
- For query
Constraints:
- The length of
wordsand the number of queries can be large, so it’s important to find an efficient solution.
Hints for Solving the Problem
-
Hint 1:
Create a helper function (or inline logic) that determines if a given string is a vowel string by checking whether its first and last characters are vowels. -
Hint 2:
Instead of checking every word for every query (which could be inefficient), precompute an auxiliary array (a prefix sum array) where each element at indexirepresents the count of vowel strings from the beginning up to indexi-1. This allows you to answer each query in constant time.
Approaches
Approach 1: Brute Force for Each Query
Explanation:
For each query [l, r], iterate over the subarray from index l to r and check if each word is a vowel string.
- Time Complexity: O(Q * k), where Q is the number of queries and k is the average length of each query range.
- Pros: Simple to implement.
- Cons: May be inefficient if there are many queries or large ranges.
Python Code (Brute Force):
Java Code (Brute Force):
Approach 2: Prefix Sum Optimization
Explanation:
Instead of processing each query separately, precompute a prefix sum array:
- Step 1: Create an auxiliary array
prefixof lengthn+1(wherenis the number of words). - Step 2: For each word, check if it is a vowel string. If yes, set
prefix[i+1] = prefix[i] + 1; otherwise, setprefix[i+1] = prefix[i]. - Step 3: For any query
[l, r], the count of vowel strings in that range is simplyprefix[r+1] - prefix[l].
Time Complexity:
- Preprocessing: O(n)
- Each query: O(1)
- Overall: O(n + Q)
Python Code (Prefix Sum):
Java Code (Prefix Sum):
Complexity Analysis
-
Brute Force Approach:
- Time Complexity: O(Q * k), where Q is the number of queries and k is the average size of the query range.
- Space Complexity: O(1) (aside from input and output storage).
-
Prefix Sum Approach:
- Time Complexity: O(n + Q) (n for building the prefix array and Q for answering queries).
- Space Complexity: O(n) (for storing the prefix sum array).
Common Mistakes & Edge Cases
-
Edge Cases:
- When
wordsis empty orqueriesis empty. - When a word has only one character (it must be a vowel to be counted).
- When no word in the given range qualifies as a vowel string.
- When
-
Common Mistakes:
- Not handling the inclusive nature of the range correctly (remember to use
r + 1when accessing the prefix sum array). - Forgetting to trim or check the first/last character’s case if the problem involves mixed-case letters (here, the problem assumes lowercase letters).
- Not handling the inclusive nature of the range correctly (remember to use
Related Problems for Further Practice
- Range Sum Query – Immutable: Precompute a prefix sum for fast range sum queries.
- Count Subarrays with Given Sum: Use prefix sums to determine subarray counts that meet certain criteria.
- Find the Longest Substring Containing Vowels: Similar string processing challenges.
GET YOUR FREE
Coding Questions Catalog
