---
description: >-
  https://leetcode.com/explore/challenge/card/30-day-leetcoding-challenge/528/week-1/3289/
---

# Counting Elements

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`

对给定数组要求元素值比它大一的元素也存在于数组中，统计满足条件的元素个数。检查是否存在用hash set。

```python
class Solution:
    def countElements(self, arr: List[int]) -> int:
        s = set(arr)
        return sum(1 if (v + 1) in s else 0 for v in arr)
    
```
