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Java Array Exercises: Separate 0s on left side and 1s on right side of an array of 0s and 1s in random order

Java Array: Exercise-51 with Solution

Write a Java program to separate 0s on left side and 1s on right side of an array of 0s and 1s in random order.

Pictorial Presentation:

Java Array Exercises: Separate 0s on left side and 1s on right side of an array of 0s and 1s in random order

Sample Solution:

Java Code:

import java.util.Arrays;

public class Main {
    public static void main(String[] args)
    {
        int arr[] = new int[]{ 0, 0, 1, 1, 0, 1, 1, 1,0 };
        int result[];
        System.out.println("Original Array ");
        System.out.println(Arrays.toString(arr));

        int n = arr.length;
 
        result = separate_0_1(arr, n);
        System.out.println("New Array ");
        System.out.println(Arrays.toString(result));
      }
    
    static int [] separate_0_1(int arr[], int n)
     {
        int count = 0;   
     
        for (int i = 0; i < n; i++) {
            if (arr[i] == 0)
                count++;
        }
 
        for (int i = 0; i < count; i++)
            arr[i] = 0;
 
        for (int i = count; i < n; i++)
            arr[i] = 1;
    
         return arr;
     }       
   }

Sample Output:

                                                                              
Original Array 
[0, 0, 1, 1, 0, 1, 1, 1, 0]
New Array 
[0, 0, 0, 0, 1, 1, 1, 1, 1]

Flowchart:

Flowchart: Separate 0s on left side and 1s on right side of an array of 0s and 1s in random order

Visualize Java code execution (Python Tutor):


Java Code Editor:

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Next: Write a Java program to separate even and odd numbers of a given array of integers. Put all even numbers first, and then odd numbers.

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Java: Tips of the Day

Different between parseInt() and valueOf() in java?

Well, the API for Integer.valueOf(String) does indeed say that the String is interpreted exactly as if it were given to Integer.parseInt(String). However, valueOf(String) returns a new Integer() object whereas parseInt(String) returns a primitive int.

If you want to enjoy the potential caching benefits of Integer.valueOf(int), you could also use this eyesore:

Integer k = Integer.valueOf(Integer.parseInt("123"))

Now, if what you want is the object and not the primitive, then using valueOf(String) may be more attractive than making a new object out of parseInt(String) because the former is consistently present across Integer, Long, Double, etc.

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