Welcome to Subscribe On Youtube

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.

Related Topics:
Array, Sort

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)

All Problems

All Solutions