Formatted question description: https://leetcode.ca/all/2148.html

# 2148. Count Elements With Strictly Smaller and Greater Elements (Easy)

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

Similar Questions:

// OJ: https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/
// Time: O(NlogU) where U is the number of unique elements in A
// Space: O(U)
class Solution {
public:
int countElements(vector<int>& A) {
set<int> s(begin(A), end(A));
int ans = 0;
for (int n : A) {
ans += s.lower_bound(n) != s.begin() && s.upper_bound(n) != s.end();
}
return ans;
}
};


Or

// OJ: https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int countElements(vector<int>& A) {
sort(begin(A), end(A));
int ans = 0;
for (int n : A) {
ans += lower_bound(begin(A), end(A), n) != begin(A) && upper_bound(begin(A), end(A), n) != end(A);
}
return ans;
}
};


## Solution 2. Exclude Smallest and Greatest Elements

// OJ: https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/
// Time: O(NlogN)
// Space: O(1)
class Solution {
public:
int countElements(vector<int>& A) {
sort(begin(A), end(A));
int i = 0, j = A.size() - 1;
while (i <= j && A[i] == A[0]) ++i;
while (i <= j && A[j] == A.back()) --j;
return j - i + 1;
}
};


Or

// OJ: https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/
// Time: O(N)
// Space: O(1)
class Solution {
public:
int countElements(vector<int>& A) {
int mn = A[0], mx = A[0], ans = A.size(), minCnt = 0, maxCnt = 0;
for (int n : A) {
if (n < mn) {
minCnt = 1;
mn = n;
} else if (n == mn) ++minCnt;
if (n > mx) {
maxCnt = 1;
mx = n;
} else if (n == mx) ++maxCnt;
}
return max(0, (int)A.size() - minCnt - maxCnt);
}
};


## Discuss

https://leetcode.com/problems/count-elements-with-strictly-smaller-and-greater-elements/discuss/1711250/C%2B%2B-Binary-Search-O(NlogN)