Counting Elements

Description

Given an integer array arr, count element x 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.

Constraints:

• 1 <= arr.length <= 1000

• 0 <= arr[i] <= 1000

Code

class Solution {
    public int countElements(int[] arr) {
        int count = 0;
        Map<Integer, Integer> map = new HashMap<>();
        for (int i = 0; i < arr.length; i++) {
            map.put(arr[i], map.getOrDefault(arr[i], 0) + 1);
        }
        Iterator<Map.Entry<Integer, Integer>> iterator = map.entrySet().iterator();
        while (iterator.hasNext()) {
            Map.Entry<Integer, Integer> next = iterator.next();
            if (map.containsKey(next.getKey() + 1)) {
                count += next.getValue();
            }
        }
        return count;
    }
}