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Formatted question description: https://leetcode.ca/all/1491.html
1491. Average Salary Excluding the Minimum and Maximum Salary (Easy)
Given an array of unique integers salary
where salary[i]
is the salary of the employee i
.
Return the average salary of employees excluding the minimum and maximum salary.
Example 1:
Input: salary = [4000,3000,1000,2000] Output: 2500.00000 Explanation: Minimum salary and maximum salary are 1000 and 4000 respectively. Average salary excluding minimum and maximum salary is (2000+3000)/2= 2500
Example 2:
Input: salary = [1000,2000,3000] Output: 2000.00000 Explanation: Minimum salary and maximum salary are 1000 and 3000 respectively. Average salary excluding minimum and maximum salary is (2000)/1= 2000
Example 3:
Input: salary = [6000,5000,4000,3000,2000,1000] Output: 3500.00000
Example 4:
Input: salary = [8000,9000,2000,3000,6000,1000] Output: 4750.00000
Constraints:
3 <= salary.length <= 100
10^3 <= salary[i] <= 10^6
salary[i]
is unique.- Answers within
10^-5
of the actual value will be accepted as correct.
Solution 1.
// OJ: https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/
// Time: O(N)
// Space: O(1)
class Solution {
public:
double average(vector<int>& A) {
return (accumulate(begin(A), end(A), 0.) - *max_element(begin(A), end(A)) - *min_element(begin(A), end(A))) / (A.size() - 2);
}
};
Java
-
class Solution { public double average(int[] salary) { int length = salary.length; double sum = 0; int max = Integer.MIN_VALUE, min = Integer.MAX_VALUE; for (int num : salary) { sum += num; max = Math.max(max, num); min = Math.min(min, num); } return (sum - max - min) / (length - 2); } }
-
// OJ: https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/ // Time: O(N) // Space: O(1) class Solution { public: double average(vector<int>& A) { return (accumulate(begin(A), end(A), 0.) - *max_element(begin(A), end(A)) - *min_element(begin(A), end(A))) / (A.size() - 2); } };
-
class Solution: def average(self, salary: List[int]) -> float: s = sum(salary) - min(salary) - max(salary) return s / (len(salary) - 2)