2148. Count Elements With Strictly Smaller and Greater Elements
Description
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
Solutions
Solution 1: Find Minimum and Maximum Values
According to the problem description, we can first find the minimum value $\textit{mi}$ and the maximum value $\textit{mx}$ of the array $\textit{nums}$. Then, traverse the array $\textit{nums}$ and count the number of elements that satisfy $\textit{mi} < x < \textit{mx}$.
The time complexity is $O(n)$, where $n$ is the length of the array $\textit{nums}$. The space complexity is $O(1)$.
Python3
class Solution:
def countElements(self, nums: List[int]) -> int:
mi, mx = min(nums), max(nums)
return sum(mi < x < mx for x in nums)
Java
class Solution {
public int countElements(int[] nums) {
int mi = Arrays.stream(nums).min().getAsInt();
int mx = Arrays.stream(nums).max().getAsInt();
int ans = 0;
for (int x : nums) {
if (mi < x && x < mx) {
ans++;
}
}
return ans;
}
}
C++
class Solution {
public:
int countElements(vector<int>& nums) {
auto [mi, mx] = ranges::minmax_element(nums);
return ranges::count_if(nums, [mi, mx](int x) { return *mi < x && x < *mx; });
}
};
Go
func countElements(nums []int) (ans int) {
mi := slices.Min(nums)
mx := slices.Max(nums)
for _, x := range nums {
if mi < x && x < mx {
ans++
}
}
return
}
TypeScript
function countElements(nums: number[]): number {
const mi = Math.min(...nums);
const mx = Math.max(...nums);
return nums.filter(x => mi < x && x < mx).length;
}