# 【LeetCode】 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.

> Constraints:
> * `1 <= arr.length <= 1000`
> * `0 <= arr[i] <= 1000`

> 給予一正整數陣列`arr`，計算存在多少元素`x`當`x + 1`也在`arr`當中。

> 如果有重複的數在`arr`當中，將它們分開計算。

> 限制：
> * `1 <= arr.length <= 1000`
> * `0 <= arr[i] <= 1000`

## Example:

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.

## Solution

* 先跑一次迴圈，使用set紀錄哪些數字存在於陣列中。
* 再跑一次迴圈，根據剛剛的set計算`count`。

### Code

class Solution {
public:
    int countElements(vector<int>& arr) {
        set<int> hash;
        for(int i = 0; i < arr.size(); i++)
            hash.insert(arr[i] - 1);
        int count = 0;
        for(int i = 0; i < arr.size(); i++)
            if(hash.find(arr[i]) != hash.end())
                count++;
        return count;
    }
};

###### tags: `LeetCode` `C++`