carloscn
commented
1 year ago Analysis
pub fn count_elements(nums: Vec<i32>) -> i32
{
let mut ret:i32 = 0;
if nums.len() < 1 {
return ret;
}
let mut max_val:i32 = i32::MIN;
let mut min_val:i32 = i32::MAX;
for i in 0..nums.len() {
max_val = max_val.max(nums[i]);
min_val = min_val.min(nums[i]);
}
for i in 0..nums.len() {
if nums[i] > min_val && nums[i] < max_val {
ret += 1;
}
}
return ret;
}
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