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Java Array Exercises: Segregate all 0s on left side and all 1s on right side of a given array of 0s and 1s

Java Array: Exercise-42 with Solution

Write a Java program to segregate all 0s on left side and all 1s on right side of a given array of 0s and 1s.

Pictorial Presentation:

Java Array Exercises: Segregate all 0s on left side and all 1s on right side of a given array of 0s and 1s

Sample Solution:

Java Code:

import java.util.*;
import java.lang.*;
public class Main
{
   public static void main (String[] args) 
    {  
        int nums[] = {0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1};
        int i,  nums_size = nums.length;
        int left = 0, right = nums_size - 1;
        
        System.out.println("Original Array : "+Arrays.toString(nums));  
 
        while (left < right) 
        {
            /* While  0 at left increment left index  */
            while (nums[left] == 0 && left < right)
               left++;
 
            /* While we see 1 at right decrement right index*/
            while (nums[right] == 1 && left < right)
                right--;
 
           
            if (left < right) 
            {
                nums[left] = 0;
                nums[right] = 1;
                left++;
                right--;
            }
        }
        
       System.out.println("Array after segregation is : "+Arrays.toString(nums));  
    }
}

Sample Output:

                                                                              
Original Array : [0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 1]
Array after segregation is : [0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1]

Flowchart:

Flowchart: Segregate all 0s on left side and all 1s on right side of a given array of 0s and 1s

Visualize Java code execution (Python Tutor):


Java Code Editor:

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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