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Java Array Exercises: Remove the duplicate elements of a given array and return the new length of the array

Java Array: Exercise-33 with Solution

Write a Java program to remove the duplicate elements of a given array and return the new length of the array.
Sample array: [20, 20, 30, 40, 50, 50, 50]
After removing the duplicate elements the program should return 4 as the new length of the array.

Pictorial Presentation:

Java Array Exercises: Remove the duplicate elements of a given array and return the new length of the array

Sample Solution:

Java Code:

public class Exercise33 {    
   public static void main(String[] args) {
        int nums[] = {20, 20, 30, 40, 50, 50, 50};  
		System.out.println("Original array length: "+nums.length);
		System.out.print("Array elements are: ");
       for (int i = 0; i < nums.length; i++)
        {
            System.out.print(nums[i]+" ");
        }
		System.out.println("\nThe new length of the array is: "+array_sort(nums));
			
    }
    
    public static int array_sort(int[] nums) {
         int index = 1;
        for (int i = 1; i < nums.length; i++) {
            if (nums[i] != nums[index-1])
                nums[index++] = nums[i];
        }
	  return index;
    }
}

Sample Output:

                                                                              
Original array length: 7                                               
Array elements are: 20 20 30 40 50 50 50                               
The new length of the array is: 4

Flowchart:

Flowchart: Java exercises: Remove the duplicate elements of a given array and return the new length of the array

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Java Code Editor:

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Next: Write a Java program to find the length of the longest consecutive elements sequence from a given unsorted array of integers.

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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