Your implementation, to me at least, doesn't seem correct. You need to remove adjacent duplicates, but looking at your last example, the sequence acaaabbacdddd completely removes the b and d characters from the slice. You're also using an awful lot of code to do a simple thing. What I'd do is quite simply this:

• Iterate over the slice from index 0 to the next to last character
• For each character, iterate over the remainder of the slice (nested loop) until you find a character that doesn't equal the current index
• For each character at the current position + 1 that matches the current one, remove it, as it's an adjacent duplicate.

The code itself is quite simple:

func dedup(s []string) []string {
    // iterate over all characters in slice except the last one
    for i := 0; i < len(s)-1;i++ {
        // iterate over all the remaining characeters
        for j := i+1; j < len(s); j++ {
            if s[i] != s[j] {
                break // this wasn't a duplicate, move on to the next char
            }
            // we found a duplicate!
            s = append(s[:i], s[j:]...)
        }
    }
    return s
}

Given an input like [g e e k s f o r g e e g], the output of this is [g e k s f o r g e g]

Demo

The only trickery here is this line: s = append(s[:i], s[j:]...). What this effectively does is reassign the slice s to contain i values starting at 0 (so if i is 2, the slice will be [g, e]). The second part is creating a slice starting at j, until the end of s. Again, if j is 2, this slice will be all values starting at offset 2 until the end ([e k s f o r g e e g]).

So let's look at an actual example:

• i == 1
• j == i+1 (2)
• s[i] == e, s[j] == e`

We have a duplicate, so we'll reassign s like so:

 s = append(s[:1], s[2:]...)`

This means we're appending [e k s f o r g e e g] to [g], removing the duplicate e. Job done.