values - write a java program to test if an array contains a specific value.

How to determine whether an array contains a particular value in Java? (17)

I have a String[] with values like so:

public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};

Given String s, is there a good way of testing whether VALUES contains s?

  1. For arrays of limited length use the following (as given by camickr). This is slow for repeated checks, especially for longer arrays (linear search).

  2. For fast performance if you repeatedly check against a larger set of elements

    • An array is the wrong structure. Use a TreeSet and add each element to it. It sorts elements and has a fast exist() method (binary search).

    • If the elements implement Comparable & you want the TreeSet sorted accordingly:

      ElementClass.compareTo() method must be compatable with ElementClass.equals(): see Triads not showing up to fight? (Java Set missing an item)

      TreeSet myElements = new TreeSet();
      // Do this for each element (implementing *Comparable*)
      // *Alternatively*, if an array is forceably provided from other code:
    • Otherwise, use your own Comparator:

      class MyComparator implements Comparator<ElementClass> {
           int compareTo(ElementClass element1; ElementClass element2) {
                // Your comparison of elements
                // Should be consistent with object equality
           boolean equals(Object otherComparator) {
                // Your equality of comparators
      // construct TreeSet with the comparator
      TreeSet myElements = new TreeSet(new MyComparator());
      // Do this for each element (implementing *Comparable*)
    • The payoff: check existence of some element:

      // Fast binary search through sorted elements (performance ~ log(size)):
      boolean containsElement = myElements.exists(someElement);

Four Different Ways to Check If an Array Contains a Value

1) Using List:

public static boolean useList(String[] arr, String targetValue) {
    return Arrays.asList(arr).contains(targetValue);

2) Using Set:

public static boolean useSet(String[] arr, String targetValue) {
    Set<String> set = new HashSet<String>(Arrays.asList(arr));
    return set.contains(targetValue);

3) Using a simple loop:

public static boolean useLoop(String[] arr, String targetValue) {
    for (String s: arr) {
        if (s.equals(targetValue))
            return true;
    return false;

4) Using Arrays.binarySearch():

The code below is wrong, it is listed here for completeness. binarySearch() can ONLY be used on sorted arrays. You will find the result is weird below. This is the best option when array is sorted.

public static boolean binarySearch(String[] arr, String targetValue) {  
            int a = Arrays.binarySearch(arr, targetValue);
            return a > 0;

Quick Example:

String testValue="test";
String newValueNotInList="newValue";
String[] valueArray = { "this", "is", "java" , "test" };
Arrays.asList(valueArray).contains(testValue); // returns true
Arrays.asList(valueArray).contains(newValueNotInList); // returns false

Arrays.asList() -> then calling the contains() method will always work, but a search algorithm is much better since you don't need to create a lightweight list wrapper around the array, which is what Arrays.asList() does.

public boolean findString(String[] strings, String desired){
   for (String str : strings){
       if (desired.equals(str)) {
           return true;
   return false; //if we get here… there is no desired String, return false.

Check this

String[] VALUES = new String[] {"AB","BC","CD","AE"};
String s;

for(int i=0; i< VALUES.length ; i++)
    if ( VALUES[i].equals(s) )
        // do your stuff
        //do your stuff

Developers often do:

Set<String> set = new HashSet<String>(Arrays.asList(arr));
return set.contains(targetValue);

The above code works, but there is no need to convert a list to set first. Converting a list to a set requires extra time. It can as simple as:



   for(String s: arr){
            return true;

return false;

The first one is more readable than the second one.

For what its worth I ran a test comparing the 3 suggestions for speed. I generated random integers, converted them to a String and added them to an array. I then searched for the highest possible number/string, which would be a worst case scenario for the asList().contains().

When using a 10K array size the results where:

Sort & Search   : 15
Binary Search   : 0
asList.contains : 0

When using a 100K array the results where:

Sort & Search   : 156
Binary Search   : 0
asList.contains : 32

So if the array is created in sorted order the binary search is the fastest, otherwise the asList().contains would be the way to go. If you have many searches, then it may be worthwhile to sort the array so you can use the binary search. It all depends on your application.

I would think those are the results most people would expect. Here is the test code:

import java.util.*;

public class Test
    public static void main(String args[])
        long start = 0;
        int size = 100000;
        String[] strings = new String[size];
        Random random = new Random();

        for (int i = 0; i < size; i++)
            strings[i] = "" + random.nextInt( size );

        start = System.currentTimeMillis();
        System.out.println(Arrays.binarySearch(strings, "" + (size - 1) ));
        System.out.println("Sort & Search : " + (System.currentTimeMillis() - start));

        start = System.currentTimeMillis();
        System.out.println(Arrays.binarySearch(strings, "" + (size - 1) ));
        System.out.println("Search        : " + (System.currentTimeMillis() - start));

        start = System.currentTimeMillis();
        System.out.println(Arrays.asList(strings).contains( "" + (size - 1) ));
        System.out.println("Contains      : " + (System.currentTimeMillis() - start));

I'm surprised no one suggested to just simply implement it by hand:

public static <T> boolean contains(final T[] array, final T v) {
    for (final T e : array)
        if (e == v || v != null && v.equals(e))
            return true;

    return false;


The v != null condition is constant inside the method, it always evaluates to the same boolean value during the method call. So if the input array is big, it is more efficient to evaluate this condition only once and we can use a simplified/faster condition inside the for loop based on the result. The improved contains() method:

public static <T> boolean contains2(final T[] array, final T v) {
    if (v == null) {
        for (final T e : array)
            if (e == null)
                return true;
    } else {
        for (final T e : array)
            if (e == v || v.equals(e))
                return true;

    return false;

If the array is not sorted, you will have to iterate over everything and make a call to equals on each.

If the array is sorted, you can do a binary search, there's one in the Arrays class.

Generally speaking, if you are going to do a lot of membership checks, you may want to store everything in a Set, not in an array.

If you have the google collections library, Tom's answer can be simplified a lot by using ImmutableSet (

This really removes a lot of clutter from the initialization proposed

private static final Set<String> VALUES =  ImmutableSet.of("AB","BC","CD","AE");

In Java 8 use Streams.

List<String> myList =
Arrays.asList("a1", "a2", "b1", "c2", "c1");

.filter(s -> s.startsWith("c"))

Just to clear the code up to start with. We have (corrected):

public static final String[] VALUES = new String[] {"AB","BC","CD","AE"};

This is a mutable static which FindBugs will tell you is very naughty. It should be private:

private static final String[] VALUES = new String[] {"AB","BC","CD","AE"};

(Note, you can actually drop the new String[]; bit.)

So, reference arrays are bad, and in particular here we want a set:

private static final Set<String> VALUES = new HashSet<String>(Arrays.asList(
     new String[] {"AB","BC","CD","AE"}

(Paranoid people, such as myself, may feel more at ease if this was wrapped in Collections.unmodifiableSet - it could even be made public.)

"Given String s, is there a good way of testing whether VALUES contains s?"



ObStupidAnswer (but I think there's a lesson in here somewhere):

enum Values {
    AB, BC, CD, AE

try {
    return true;
} catch (IllegalArgumentException exc) {
    return false;

Try this:

ArrayList<Integer> arrlist = new ArrayList<Integer>(8);

// use add() method to add elements in the list

boolean retval = arrlist.contains(10);
if (retval == true) {
    System.out.println("10 is contained in the list");
else {
    System.out.println("10 is not contained in the list");

Use Array.BinarySearch(array,obj) for finding the given object in array or not. Ex:

if (Array.BinarySearch(str, i) > -1) -->true --exists

false --not exists

With Java 8 you can create a stream and check if any entries in the stream matches "s":

String[] values = {"AB","BC","CD","AE"};
boolean sInArray ="s"::equals);

Or as a generic method:

public static <T> boolean arrayContains(T[] array, T value) {


Warning: this doesn't work for arrays of primitives (see the comments).

Since java-8 you can now use Streams.

String[] values = {"AB","BC","CD","AE"};
boolean contains ="s"::equals);

To check whether an array of int, double or long contains a value use IntStream, DoubleStream or LongStream respectively.


int[] a = {1,2,3,4};
boolean contains = IntStream.of(a).anyMatch(x -> x == 4);