Formatted question description: https://leetcode.ca/all/1426.html

# 1426. Counting Elements

Easy

## Description

Given an integer array arr, count how many elements x there are, such that x + 1 is also in arr.

If there’re duplicates in arr, count them seperately.

Example 1:

Input: arr = [1,2,3]

Output: 2

Explanation: 1 and 2 are counted cause 2 and 3 are in arr.

Example 2:

Input: arr = [1,1,3,3,5,5,7,7]

Output: 0

Explanation: No numbers are counted, cause there’s no 2, 4, 6, or 8 in arr.

Example 3:

Input: arr = [1,3,2,3,5,0]

Output: 3

Explanation: 0, 1 and 2 are counted cause 1, 2 and 3 are in arr.

Example 4:

Input: arr = [1,1,2,2]

Output: 2

Explanation: Two 1s are counted cause 2 is in arr.

Example 5:

Input: arr = [1,1,2]

Output: 2

Explanation: Both 1s are counted because 2 is in the array.

Constraints:

• 1 <= arr.length <= 1000
• 0 <= arr[i] <= 1000

## Solution

Use a set to store all the elements in arr. Then loop over arr. For each element x in arr, if x + 1 is in the set, then increase the counter by 1. Finally, return the counter.

class Solution {
public int countElements(int[] arr) {
Set<Integer> set = new HashSet<Integer>();
for (int num : arr)