Check whether a string slice contains a certain value in Go

If you have a slice of strings in an arbitrary order, finding if a value exists in the slice requires O(n) time. This applies to all languages.

If you intend to do a search over and over again, you can use other data structures to make lookups faster. However, building these structures require at least O(n) time. So you will only get benefits if you do lookups using the data structure more than once.

For example, you could load your strings into a map. Then lookups would take O(1) time. Insertions also take O(1) time making the initial build take O(n) time:

set := make(map[string]bool)
for _, v := range list {
    set[v] = true
}

fmt.Println(set["b"])

You can also sort your string slice and then do a binary search. Binary searches occur in O(log(n)) time. Building can take O(n*log(n)) time.

sort.Strings(list)
i := sort.SearchStrings(list, "b")
fmt.Println(i < len(list) && list[i] == "b")

Although in theory given an infinite number of values, a map is faster, in practice it is very likely searching a sorted list will be faster. You need to benchmark it yourself.

You can use a map, and have the value e.g. a bool

m := map[string] bool {"a":true, "b":true, "x":true}
if m["a"] { // will be false if "a" is not in the map
    //it was in the map
}

There's also the sort package, so you could sort and binary search your slices

To replace sets you should use a map[string]struct{}. This is efficient and considered idiomatic, the "values" take absolutely no space.

Initialize the set:

set := make(map[string]struct{})

Put an item :

set["item"]=struct{}{}

Check whether an item is present:

_, isPresent := set["item"]

Remove an item:

delete(set, "item")