#### QUESTION:

Given an integer array nums, return the number of elements that have both a strictly smaller and a strictly greater element appear in nums.

**Example 1:**

```
Input: nums = [11,7,2,15]
Output: 2
Explanation: The element 7 has the element 2 strictly smaller than it and the element 11 strictly greater than it.
Element 11 has element 7 strictly smaller than it and element 15 strictly greater than it.
In total there are 2 elements having both a strictly smaller and a strictly greater element appear in nums.
```

**Example 2:**

```
Input: nums = [-3,3,3,90]
Output: 2
Explanation: The element 3 has the element -3 strictly smaller than it and the element 90 strictly greater than it.
Since there are two elements with the value 3, in total there are 2 elements having both a strictly smaller and a strictly greater element appear in nums.
```

**Constraints:**

```
1 <= nums.length <= 100
-105 <= nums[i] <= 105
```

#### EXPLANATION:

如果当前数据只有2种以下, 那么久返回0 , 否则就直接减去min的count和max的count即可.

#### SOLUTION:

```
class Solution {
func countElements(_ nums: [Int]) -> Int {
return Set(nums).count <= 2 ? 0 : nums.count - nums.filter({ a in
a == nums.min()
}).count - nums.filter({ b in
b == nums.max()
}).count
}
}
```