Leetcode 1119

The first approach is pretty simple. If the letter is not a vowel, add it to a resulting string.

The second approach is better because we keep on using the same string. We remake the string every time there is a vowel by going around the vowel.

This can be shown with some images:

We can have:

input: "leetcode"
expected output: "ltcd"

We can fast forward to our first vowel, 'e'.

Now we re-assign our string s to "leetcode" without the 'e' and we get:

We have circled the letter 't' because i == 2 but we forgot about the second 'e' so we subtract 1 from i and we get:

As we can see, 'e' is a vowel, so we have to remake the string to:

Note: As you can see, we skipped a lot of the letter because it would take a lot longer if we didn’t ignore them
You can see that the 'o' is a vowel, so we can remake the string to:

We skipped some more letters again because the other letters were not needed.
The 'e' is a vowel, so we can remake the string to:

As we can see, ltdc is equal to our expected output.

The First Code:

func removeVowels(s string) string {
    res := ""
    for _, i2 := range s {
        if i2 != 'a' && i2 != 'e' && i2 != 'i' && i2 != 'o' && i2 != 'u' {
            res += string(i2)
        }
    }
    return res
}

func removeVowels(s string) string {
    for i := 0; i < len(s); i++ {
        if s[i] == 'a' || s[i] == 'e' || s[i] == 'i' || s[i] == 'o' || s[i] == 'u' {
            s = s[:i] + s[i+1:]
            i -= 1
        }
    }
    return s
}