BigBinary Academy

DSA using JavaScript

Lesson

8.5Count Vowels

Problem Statement

Count the number of vowels (a, e, i, o, u) in a given string.

Example Inputs

Input: "hello"
- Output: 2
Input: "JavaScript"
- Output: 3
Input: "java"
- Output: 2
Input: "Education"
- Output: 5

JavaScript Method - Using Only String Methods

Approach

To count the vowels in a string, we can convert the string to lowercase to handle case insensitivity, then use the .split() and .filter() methods to isolate vowels and count them.

Steps

Convert the string to lowercase using .toLowerCase().
Split the string into an array of characters with .split('').
Use .filter() to create a new array containing only vowels (a, e, i, o, u).
Return the length of the array to get the count of vowels.

Time & Space Complexity

Time Complexity: O(n), where n is the length of the string, as we iterate over each character once.
Space Complexity: O(n), due to the creation of a new array for filtered vowels.

Code Snippet

Dry Run

Example input: "hello"

Step 1: str.toLowerCase() → "hello"
Step 2: "hello".split('') → ['h', 'e', 'l', 'l', 'o']
Step 3: filter(['h', 'e', 'l', 'l', 'o']) → ['e', 'o']
Step 4: Return ['e', 'o'].length → 2

Alternative Approach - Without Using JavaScript String Methods

Approach

In this approach, we loop through the string and manually count vowels by checking each character. This avoids using JavaScript string methods directly.

Steps

Initialize a variable count to zero.
Loop through each character of the string.
Convert the character to lowercase (using conditional checks instead of methods).
Check if the character is a vowel (a, e, i, o, u).
If it is, increment the count variable by one.
Return the count after the loop ends.

Time & Space Complexity

Time Complexity: O(n), as we traverse each character once.
Space Complexity: O(1), as we only use a counter variable.

Code Snippet

Dry Run

Example input: "hello"

Initialize: count = 0
Loop through "hello":
- h → Not a vowel, count = 0
- e → Vowel, count = 1
- l → Not a vowel, count = 1
- l → Not a vowel, count = 1
- o → Vowel, count = 2
Final output: 2

Complexity Analysis

JavaScript Method (Using String Methods)

Time Complexity: O(n)
Space Complexity: O(n)

Alternative Approach (Without String Methods)

Time Complexity: O(n)
Space Complexity: O(1)

Conclusion

Counting vowels in a string can be done using string methods like split and filter for simplicity or with a loop for efficiency. Both methods have similar time complexities, but the manual loop approach saves space by avoiding additional arrays.