How do you solve Count Vowel Strings in Ranges on LeetCode?

Count Vowel Strings in Ranges (LeetCode #2559) can be solved using Brute Force. The optimal approach is Brute Force with O(n²) time complexity and O(1) space complexity. Key insight: The brute force approach checks each string in the specified range for every query. This is straightforward but inefficient, especially for large inputs.

What is the time complexity of Count Vowel Strings in Ranges?

The optimal solution for Count Vowel Strings in Ranges runs in O(n²) time and uses O(1) space. The time complexity is O(n²) because for each query, we may check up to n strings, leading to a quadratic number of checks in the worst case.

What algorithm or data structure is used to solve Count Vowel Strings in Ranges?

Count Vowel Strings in Ranges uses the following concepts: Array, String, Prefix Sum. The recommended patterns to study are: .

Count Vowel Strings in Ranges — LeetCode #2559 (Medium)

Tags: Array, String, Prefix Sum

Related patterns:

Brute Force approach

Time complexity: O(n²). Space complexity: O(1).

The brute force approach checks each string in the specified range for every query. This is straightforward but inefficient, especially for large inputs.

The time complexity is O(n²) because for each query, we may check up to n strings, leading to a quadratic number of checks in the worst case.

Step 1: For each query, extract the range [l, r].
Step 2: Loop through the strings in the range from l to r.
Step 3: For each string, check if it starts and ends with a vowel.
Step 4: Count the number of strings that satisfy the condition.
Step 5: Store the count for the current query.

For words = ["aba", "bcb", "ece", "aa", "e"] and queries = [[0, 2], [1, 4], [1, 1]]: 1. Query [0, 2]: Check words[0] = 'aba' (valid), words[1] = 'bcb' (invalid), words[2] = 'ece' (valid) => Count = 2 2. Query [1, 4]: Check words[1] = 'bcb' (invalid), words[2] = 'ece' (valid), words[3] = 'aa' (valid), words[4] = 'e' (valid) => Count = 3 3. Query [1, 1]: Check words[1] = 'bcb' (invalid) => Count = 0 Final output: [2, 3, 0]

#2559

Count Vowel Strings in Ranges

Medium

Array↗String↗Prefix Sum↗

LeetCode ↗

Approaches

💡

Intuition

Time O(n²)Space O(1)

The brute force approach checks each string in the specified range for every query. This is straightforward but inefficient, especially for large inputs.

⚙️

Algorithm

5 steps

1Step 1: For each query, extract the range [l, r].
2Step 2: Loop through the strings in the range from l to r.
3Step 3: For each string, check if it starts and ends with a vowel.
4Step 4: Count the number of strings that satisfy the condition.
5Step 5: Store the count for the current query.

solution.py13 lines

1# Full working Python code
2words = ["aba", "bcb", "ece", "aa", "e"]
3queries = [[0, 2], [1, 4], [1, 1]]
4vowels = {'a', 'e', 'i', 'o', 'u'}
5
6ans = []
7for l, r in queries:
8    count = 0
9    for i in range(l, r + 1):
10        if words[i][0] in vowels and words[i][-1] in vowels:
11            count += 1
12    ans.append(count)
13print(ans)

ℹ

Complexity note: The time complexity is O(n²) because for each query, we may check up to n strings, leading to a quadratic number of checks in the worst case.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.