How do you solve Count the Number of Vowel Strings in Range on LeetCode?

Count the Number of Vowel Strings in Range (LeetCode #2586) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Vowel checking can be efficiently done using a set for O(1) lookups.

What is the brute force solution for Count the Number of Vowel Strings in Range?

The brute force approach for Count the Number of Vowel Strings in Range is Brute Force, with O(n) time complexity and O(1) space complexity. The brute force approach checks each string in the specified range to see if it starts and ends with a vowel. This is straightforward but can be inefficient for larger inputs. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Count the Number of Vowel Strings in Range?

Count the Number of Vowel Strings in Range uses the following concepts: Array, String, Counting. The recommended patterns to study are: Hash Set, Array.

What are the interview tips for Count the Number of Vowel Strings in Range?

Always clarify the constraints and edge cases with the interviewer. Discuss your thought process while coding to demonstrate your understanding. Consider the efficiency of your solution and be prepared to explain your choices.

What are common mistakes when solving Count the Number of Vowel Strings in Range?

Not checking both the first and last characters correctly. Overcomplicating the solution with unnecessary data structures.

#2586

Count the Number of Vowel Strings in Range

Easy

ArrayStringCountingHash SetArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n)	O(n)
Space	O(1)	O(1)

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Intuition

Time O(n)Space O(1)

The optimal solution is similar to the brute force approach but focuses on reducing unnecessary checks by directly accessing the first and last characters of each string. This ensures we only loop through the relevant range once.

⚙️

Algorithm

6 steps

1Step 1: Create a set of vowels for quick lookup.
2Step 2: Initialize a counter to zero.
3Step 3: Loop through the words array from index 'left' to 'right'.
4Step 4: For each word, check if the first and last characters are vowels using the set.
5Step 5: If both conditions are met, increment the counter.
6Step 6: Return the counter.

solution.py7 lines

1def countVowelStrings(words, left, right):
2    vowels = {'a', 'e', 'i', 'o', 'u'}
3    count = 0
4    for i in range(left, right + 1):
5        if words[i][0] in vowels and words[i][-1] in vowels:
6            count += 1
7    return count

ℹ

Complexity note: The time complexity remains O(n) as we still iterate through the range once. The space complexity is O(1) since we only use a constant amount of space for the counter and the vowel set.

1Vowel checking can be efficiently done using a set for O(1) lookups.
2The problem can be solved in a single pass through the specified range.

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