README_EN.md

comments	difficulty	edit_url	rating	source	tags
true	Easy	https://github.com/doocs/leetcode/edit/main/solution/2100-2199/2148.Count%20Elements%20With%20Strictly%20Smaller%20and%20Greater%20Elements/README_EN.md	1201	Weekly Contest 277 Q1	Array Counting Sorting

2148. Count Elements With Strictly Smaller and Greater Elements

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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;
}