java - Find first matching string in two lists in less than O(n^2) time using a Java8 one-liner -

simple question, have 2 arrays / lists, i'm looking efficient one-liner in java8 match first argument against long list of possible arguments.

arguments public static void main(string... args) , other 1 list of predefined strings i'll call secondarray, ["a", "b", "cc", "de", ...]. want match 2 incoming arrays against each other , find first matching argument against secondarray list.

example: args = arrays.aslist("r", "b", "c", "d", ...)

secondlist = arrays.aslist("q", "z", "x", "c", "d", ...)

the first matching argument "c"

the java7 (inefficient) way of doing have been:

string profile = "default"; (string arg : args){     if (secondarray.contains(arg)){         profile = arg;         break;     } }  dosomethingwith(profile); 

in java8, i've done far, wondering if there's better way (more efficient, faster, etc - let's assume both args , secondarray can potentially contain 100k entries):

    string profile = arrays.stream(args)             .filter(secondarray::contains)             .findfirst()             .orelse("default");      dosomethingwith(profile); 

my assumption filter doing secondarray::contains mean o(n^2) complexity, since secondarray.contains(arg) called each of arguments. closer o(n) while still maintaining single line of java8 i'm after.

the trick make 1 liner create set in filter cause

string profile = arrays.stream(args)             .filter(new hashset(secondlist)::contains)             .findfirst()             .orelse("default"); 

such set going created once. same instance of set used each item in args.

so code o(n) create set , o(n) finding first arg.

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just keep in mind expanded lambda like

.filter(arg -> new hashset(secondlist).contains(arg)) 

will create set each argument , blowup complexity o(n^2).