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# Determine If All Elements Are the Same in a Java List

Last modified: January 19, 2022

## 1. Overview

In this quick tutorial, we’ll find out how to determine if all the elements in a *List* are the same.

We'll also look at the time complexity of each solution using Big O notation, giving us the worst case scenario.

## 2. Example

Let’s suppose we have the following 3 lists:

```
notAllEqualList = Arrays.asList("Jack", "James", "Sam", "James");
emptyList = Arrays.asList();
allEqualList = Arrays.asList("Jack", "Jack", "Jack", "Jack");
```

Our task is to propose different solutions that return *true* only for *emptyList* and *allEqualList*.

## 3. Basic Looping

First, it's true that for all elements to be equal, they all have to equal the first element. Let's take advantage of that in a loop:

```
public boolean verifyAllEqualUsingALoop(List<String> list) {
for (String s : list) {
if (!s.equals(list.get(0)))
return false;
}
return true;
}
```

This is nice because, while the time complexity is *O(n)*, it may often exit early.

## 4. *HashSet*

We can also use a *HashSet* since all its elements are distinct. I**f we convert a List to a HashSet and the resulting size is less than or equal to 1, then we know that all elements in the list are equal:**

```
public boolean verifyAllEqualUsingHashSet(List<String> list) {
return new HashSet<String>(list).size() <= 1;
}
```

Converting a *List* to *HashSet* costs *O(n)* time while calling *size* takes *O(1)*. Thus, we still have a total time complexity of *O(n)*.

## 5. *Collections* API

Another solution is to use the *frequency(Collection c, Object o)* method of the Collections API. **This method returns the number of elements in a Collection c matching an Object o**.

So, if the frequency result is equal to the size of the list, we know that all the elements are equal:

```
public boolean verifyAllEqualUsingFrequency(List<String> list) {
return list.isEmpty() || Collections.frequency(list, list.get(0)) == list.size();
}
```

Similar to the previous solutions, the time complexity is *O(n)* since internally,* Collections.frequency()* uses basic looping.

## 6. Streams

The *Stream* API in Java 8 gives us even more alternative ways of detecting whether all items in a list are equal.

### 6.1. *distinct()*

Let's look at one particular solution making use of the *distinct() *method.

To verify if all the elements in a list are equal, **we count the distinct elements of its stream:**

```
public boolean verifyAllEqualUsingStream(List<String> list) {
return list.stream()
.distinct()
.count() <= 1;
}
```

If the count of this stream is smaller or equal to 1, then all the elements are equal and we return *true*.

The total cost of the operation is *O(n),* which is the time taken to go through all the stream elements.

### 6.2. *allMatch()*

The *Stream* API's *allMatch()* method provides a perfect solution to determine whether all elements of this stream match the provided predicate:

```
public boolean verifyAllEqualAnotherUsingStream(List<String> list) {
return list.isEmpty() || list.stream()
.allMatch(list.get(0)::equals);
}
```

Similar to the previous example using streams, this one has an *O(n)* time complexity, which is the time to traverse the whole stream.

## 7. Third-Party Libraries

If we're stuck on an earlier version of Java and cannot use the Stream API, **we can make use of third-party libraries such as Google Guava and Apache Commons.**

Here, we have two solutions that are very much alike, iterating through a list of elements and matching it with the first element. Thus, we can easily calculate the time complexity to be *O(n)*.

### 7.1. Maven Dependencies

To use either, we can add either *guava* or *commons-collections4* respectively to our project:

```
<dependency>
<groupId>com.google.guava</groupId>
<artifactId>guava</artifactId>
<version>31.0.1-jre</version>
</dependency>
```

```
<dependency>
<groupId>org.apache.commons</groupId>
<artifactId>commons-collections4</artifactId>
<version>4.1</version>
</dependency>
```

### 7.2. Google Guava

In *Google Guava*, the static method *Iterables.all()* returns *true* if all elements in the list satisfy the predicate:

```
public boolean verifyAllEqualUsingGuava(List<String> list) {
return Iterables.all(list, new Predicate<String>() {
public boolean apply(String s) {
return s.equals(list.get(0));
}
});
}
```

### 7.3. Apache Commons

Similarly, the *Apache Commons* library also provides a utility class *IterableUtils* with a set of static utility methods to operate on *Iterable* instances.

In particular, the static method *IterableUtils.matchesAll()* returns *true* if all elements in the list satisfy the predicate:

```
public boolean verifyAllEqualUsingApacheCommon(List<String> list) {
return IterableUtils.matchesAll(list, new org.apache.commons.collections4.Predicate<String>() {
public boolean evaluate(String s) {
return s.equals(list.get(0));
}
});
}
```

## 8. Conclusion

In this article, we’ve learned different ways of verifying whether all elements in a *List *are equal starting with simple Java functionality and then showing alternative ways using the *Stream* API and the third-party libraries *Google Guava* and *Apache Commons.*

We have also learned that each of the solutions gives us the same time complexity of *O(n)*. However, it's up to us to choose the best one according to how and where it will be used.

And make sure to check out the complete set of samples over on GitHub.