How to check if an array (unsorted) contains a certain value? This is a very useful and frequently used operation in Java. It is also a top voted question on Stack Overflow. As shown in top voted answers, this can be done in several different ways, but the time complexity could be very different. In the following I will show the time cost of each method.

1. 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():
binarySearch() can ONLY be used on sorted arrays. If the array is sorted, you can use the following code to search the target element:

public static boolean useArraysBinarySearch(String[] arr, String targetValue) {	
	int a =  Arrays.binarySearch(arr, targetValue);
	if(a > 0)
		return true;
	else
		return false;
}

2. Time Complexity

The approximate time cost can be measured by using the following code. The basic idea is to search an array of size 5, 1k, 10k. The approach may not be precise, but the idea is clear and simple.

public static void main(String[] args) {
	String[] arr = new String[] {  "CD",  "BC", "EF", "DE", "AB"};
 
	//use list
	long startTime = System.nanoTime();
	for (int i = 0; i < 100000; i++) {
		useList(arr, "A");
	}
	long endTime = System.nanoTime();
	long duration = endTime - startTime;
	System.out.println("useList:  " + duration / 1000000);
 
	//use set
	startTime = System.nanoTime();
	for (int i = 0; i < 100000; i++) {
		useSet(arr, "A");
	}
	endTime = System.nanoTime();
	duration = endTime - startTime;
	System.out.println("useSet:  " + duration / 1000000);
 
	//use loop
	startTime = System.nanoTime();
	for (int i = 0; i < 100000; i++) {
		useLoop(arr, "A");
	}
	endTime = System.nanoTime();
	duration = endTime - startTime;
	System.out.println("useLoop:  " + duration / 1000000);
}

Result:

useList:  13
useSet:  72
useLoop:  5

Use a larger array (1k):

String[] arr = new String[1000];
 
Random s = new Random();
for(int i=0; i< 1000; i++){
	arr[i] = String.valueOf(s.nextInt());
}

Result:

useList:  112
useSet:  2055
useLoop:  99
useArrayBinary:  12

Use a larger array (10k):

String[] arr = new String[10000];
 
Random s = new Random();
for(int i=0; i< 10000; i++){
	arr[i] = String.valueOf(s.nextInt());
}

Result:

useList:  1590
useSet:  23819
useLoop:  1526
useArrayBinary:  12

Clearly, using a simple loop method is more efficient than using any collection. A lot of developers use the first method, but it is inefficient. Pushing the array to another collection requires spin through all elements to read them in before doing anything with the collection type.

The array must be sorted, if Arrays.binarySearch() method is used. In this case, the array is not sorted, therefore, it should not be used.

Actually, if you need to check if a value is contained in some array/collection efficiently, a sorted list or tree can do it in O(log(n)) or hashset can do it in O(1).

Related posts:

1. ArrayList vs. LinkedList vs. Vector
2. HashSet vs. TreeSet vs. LinkedHashSet
3. Efficient Counter in Java
4. LeetCode – Search in Rotated Sorted Array (Java)

Category >> Array >> Java If you want someone to read your code, please put the code inside <pre><code> and </code></pre> tags. For example:

<pre><code> 
String foo = "bar";
</code></pre>

• Brandy Knapp Smallwood

  try this for(x=0;10>x;x++) { any code here}
  it will repeat

• wenchao

  hashset can do it in O(1)? Why in your case, hastset is the slowest?

• I dont know

  Is there a way to say “ignore case” while using

  Arrays.asList(arr).contains(targetValue)

• Anttix

  Profiling with simple loops does not yield meaningful results.

  JVMs are smart and can optimize out code that produces data which is not consumed.
  Also the is the state of JIT caches, the garbage collector runs and other considerations.
  The bottom line is that in order to get meaningful results for small snippets you need to use a more elaborate benchmarking harness. e.g. JMH. See this excellent talk about benchmarking below

  http://shipilev.net/talks/devoxx-Nov2013-benchmarking.pdf

How to Check if an Array Contains a Value in Java Efficiently?