Replacing vowels with their 1-based index in a string in JavaScript

We are required to write a JavaScript function that takes in a string and replaces all occurrences of the vowels in the string with their index in the string (1-based).

It means if the second letter of the string is a vowel, it should be replaced by 2.

Example

Following is the code ?

const str = 'cancotainsomevowels';
const replaceVowels = (str = '') => {
    const vowels = 'aeiou';
    let res = '';
    for(let i = 0; i 

Output

Following is the console output ?

c2nc5t78ns11m13v15w17ls


Alternative Approach Using Array Methods

We can also solve this using array methods like split() and map():

const str = 'hello world';
const replaceVowelsWithMap = (str = '') => {
    const vowels = 'aeiouAEIOU';
    return str
        .split('')
        .map((char, index) => vowels.includes(char) ? index + 1 : char)
        .join('');
};

console.log(replaceVowelsWithMap(str));
console.log(replaceVowelsWithMap('EDUCATION'));


h2ll4 w6rld
2DUC6T71012


How It Works

The function works by:

• Iterating through each character in the string
• Checking if the character is a vowel (a, e, i, o, u)
• If it's a vowel, replacing it with its 1-based index position
• If it's not a vowel, keeping the original character

Conclusion

Both approaches effectively replace vowels with their 1-based position. The loop method is more straightforward, while the array method offers a functional programming style for the same result.