# Java Array Exercises: Sort a given binary array in linear times

## Java Array: Exercise-56 with Solution

Write a Java program to sort a given binary array in linear times.

From Wikipedia,

Linear time: An algorithm is said to take linear time, or O(n) time, if its time complexity is O(n). Informally, this means that the running time increases at most linearly with the size of the input. More precisely, this means that there is a constant c such that the running time is at most cn for every input of size n. For example, a procedure that adds up all elements of a list requires time proportional to the length of the list, if the adding time is constant, or, at least, bounded by a constant.

Linear time is the best possible time complexity in situations where the algorithm has to sequentially read its entire input. Therefore, much research has been invested into discovering algorithms exhibiting linear time or, at least, nearly linear time. This research includes both software and hardware methods. There are several hardware technologies which exploit parallelism to provide this. An example is content-addressable memory. This concept of linear time is used in string matching algorithms such as the Boyer–Moore algorithm and Ukkonen's algorithm.

Example:

Input :

b_nums[] = { 0, 1, 1, 0, 1, 1, 0, 1, 0, 0 }

Output:

After sorting: [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]

**Sample Solution**:

**Java Code:**

```
import java.util.Arrays;
class solution
{
public static void sort_binary_nums(int[] b_nums)
{
int k = 0;
for (int i = 0; i < b_nums.length; i++)
{
if (b_nums[i] == 0) {
b_nums[k++] = 0;
}
}
for (int i = k; i < b_nums.length; i++) {
b_nums[k++] = 1;
}
}
public static void main (String[] args)
{
int b_nums[] = { 0, 1, 1, 0, 1, 1, 0, 1, 0, 0 };
System.out.println("Original array: "+Arrays.toString(b_nums));
sort_binary_nums(b_nums);
System.out.println("After sorting: "+Arrays.toString(b_nums));
}
}
```

Sample Output:

Original array: [0, 1, 1, 0, 1, 1, 0, 1, 0, 0] After sorting: [0, 0, 0, 0, 0, 1, 1, 1, 1, 1]

**Flowchart:**

**Java Code Editor:**

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**Previous:** Write a Java program to print all sub-arrays with 0 sum present in a given array of integers.

**Next:** Write a Java program to check if a sub-array is formed by consecutive integers from a given array of integers.

**What is the difficulty level of this exercise?**

## Java: Tips of the Day

**Array vs ArrayLists:**

The main difference between these two is that an Array is of fixed size so once you have created an Array you cannot change it but the ArrayList is not of fixed size. You can create instances of ArrayLists without specifying its size. So if you create such instances of an ArrayList without specifying its size Java will create an instance of an ArrayList of default size.

Once an ArrayList is full it re-sizes itself. In fact, an ArrayList is internally supported by an array. So when an ArrayList is resized it will slow down its performance a bit as the contents of the old Array must be copied to a new Array.

At the same time, it's compulsory to specify the size of an Array directly or indirectly while creating it. And also Arrays can store both primitives and objects while ArrayLists only can store objects.

Ref: https://bit.ly/3o8L2KH

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