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Home/ Questions/Q 8879979
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Editorial Team
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Editorial Team
Asked: June 14, 20262026-06-14T20:04:24+00:00 2026-06-14T20:04:24+00:00

What is the best way to check whether a certain value is in a

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What is the best way to check whether a certain value is in a string slice? I would use a Set in other languages, but Go doesn’t have one.

My best try is this so far:

package main

import "fmt"

func main() {
    list := []string{"a", "b", "x"}
    fmt.Println(isValueInList("b", list))
    fmt.Println(isValueInList("z", list))
}

func isValueInList(value string, list []string) bool {
    for _, v := range list {
        if v == value {
            return true
        }
    }
    return false
}

http://play.golang.org/p/gkwMz5j09n

This solution should be ok for small slices, but what to do for slices with many elements?

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  1. Editorial Team
    Editorial Team
    2026-06-14T20:04:25+00:00Added an answer on June 14, 2026 at 8:04 pm

    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.

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