LeetCode in Kotlin
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
Solution
class Solution {
fun countElements(nums: IntArray): Int {
var min = nums[0]
var max = nums[0]
var minocr = 1
var maxocr = 1
for (i in 1 until nums.size) {
if (nums[i] < min) {
min = nums[i]
minocr = 1
} else if (nums[i] == min) {
minocr++
}
if (nums[i] > max) {
max = nums[i]
maxocr = 1
} else if (nums[i] == max) {
maxocr++
}
}
return if (min == max) 0 else nums.size - minocr - maxocr
}
}