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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
-
class Solution { public int countElements(int[] nums) { int mi = 1000000, mx = -1000000; for (int num : nums) { mi = Math.min(mi, num); mx = Math.max(mx, num); } int ans = 0; for (int num : nums) { if (mi < num && num < mx) { ++ans; } } return ans; } }
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class Solution { public: int countElements(vector<int>& nums) { int mi = 1e6, mx = -1e6; for (int num : nums) { mi = min(mi, num); mx = max(mx, num); } int ans = 0; for (int num : nums) if (mi < num && num < mx) ++ans; return ans; } };
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class Solution: def countElements(self, nums: List[int]) -> int: mi, mx = min(nums), max(nums) return sum(mi < num < mx for num in nums)
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func countElements(nums []int) int { mi, mx := int(1e6), -int(1e6) for _, num := range nums { if num < mi { mi = num } if num > mx { mx = num } } ans := 0 for _, num := range nums { if mi < num && num < mx { ans++ } } return ans }
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function countElements(nums: number[]): number { const min = Math.min(...nums), max = Math.max(...nums); let ans = 0; for (let i = 0; i < nums.length; ++i) { let cur = nums[i]; if (cur < max && cur > min) { ++ans; } } return ans; }