LeetCode-in-Java
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
public class Solution {
public int countElements(int[] nums) {
int min = nums[0];
int max = nums[0];
int minocr = 1;
int maxocr = 1;
for (int i = 1; i < nums.length; ++i) {
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 min == max ? 0 : nums.length - minocr - maxocr;
}
}