Write a Java program to remove the duplicate elements of a given array and return the new length of the array

Sure, here is a simple Java program that removes duplicate elements from an array using a HashSet, which does not allow duplicate values. The program then returns the new length of the array.

import java.util.HashSet;

public class Main {
    public static void main(String[] args) {
        int[] array = {1, 2, 2, 3, 4, 4, 5, 5, 6};
        int newLength = removeDuplicates(array);
        System.out.println("The new length of the array is: " + newLength);
    }

    public static int removeDuplicates(int[] array) {
        HashSet<Integer> set = new HashSet<>();
        for (int i = 0; i < array.length; i++) {
            set.add(array[i]);
        }
        return set.size();
    }
}

In this program, we first define an array with some duplicate elements. We then call the removeDuplicates method, which takes this array as an argument. Inside this method, we create a HashSet and add all elements of the array to this set. Since a set does not allow duplicate values, all duplicates are automatically removed. Finally, we return the size of the set, which is the new length of the array without duplicates.

Sure, here is a simple Java program that removes duplicate elements from an array using a HashSet, which does not allow duplicate values. The program then returns the new length of the array.

import java.util.HashSet;

public class Main {
    public static void main(String[] args) {
        int[] array = {1, 2, 2, 3, 4, 4, 5, 5, 6};
        int newLength = removeDuplicates(array);
        System.out.println("The new length of the array is: " + newLength);
    }

    public static int removeDuplicates(int[] array) {
        HashSet<Integer> set = new HashSet<>();
        for (int i = 0; i < array.length; i++) {
            set.add(array[i]);
        }
        return set.size();
    }
}

In this program, the removeDuplicates method takes an array as input and adds each element to a HashSet. Since a HashSet does not allow duplicate values, any duplicate elements in the array are automatically removed. The method then returns the size of the HashSet, which is the new length of the array.

Here is a simple Java program that removes duplicate elements from an array and returns the new length of the array. This program uses a HashSet, which does not allow duplicate elements.

import java.util.HashSet;

public class Main {
    public static void main(String[] args) {
        int[] array = {1, 2, 2, 3, 4, 4, 5, 5, 6};
        int newLength = removeDuplicates(array);
        System.out.println("The new length of the array is: " + newLength);
    }

    public static int removeDuplicates(int[] array) {
        HashSet<Integer> set = new HashSet<Integer>();
        for (int i = 0; i < array.length; i++) {
            set.add(array[i]);
        }
        return set.size();
    }
}

In this program, the removeDuplicates method takes an array as input and adds each element to a HashSet. Since a HashSet does not allow duplicate elements, any duplicates in the array are automatically removed. The method then returns the size of the HashSet, which is the new length of the array without duplicates.

Here is a simple Java program that removes duplicate elements from an array and returns the new length of the array:

import java.util.Arrays;

public class Main {
    public static void main(String[] args) {
        int[] nums = {1, 1, 2, 2, 3, 3, 4, 4, 5, 5};
        int newLength = removeDuplicates(nums);
        System.out.println("The new length of the array is: " + newLength);
    }

    public static int removeDuplicates(int[] nums) {
        if (nums.length == 0) return 0;
        int i = 0;
        for (int j = 1; j < nums.length; j++) {
            if (nums[j] != nums[i]) {
                i++;
                nums[i] = nums[j];
            }
        }
        return i + 1;
    }
}

In this program, we first sort the array. Then we use two pointers, i and j, to traverse the array. i is the slow-runner while j is the fast-runner. As long as nums[i] == nums[j], we increment j to skip the duplicate. When we encounter nums[j] != nums[i], the duplicate run has ended so we must copy its value to nums[i + 1]. i is then incremented and we repeat the same process again until j reaches the end of array.