You are given an array of unique integers salary where salary[i] is the salary of the ith employee.

Return the average salary of employees excluding the minimum and maximum salary. Answers within 10-5 of the actual answer will be accepted.

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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


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Constraints:


	3 &lt;= salary.length &lt;= 100
	1000 &lt;= salary[i] &lt;= 106
	All the integers of salary are unique.


class Solution:
    def average(self, salary: List[int]) -> float:       

        return (sum(salary)-min(salary)-max(salary))/(len(salary)-2)
class Solution {
    public double average(int[] salary) {
        int n = salary.length-2;
        int max = Integer.MIN_VALUE;
        int min = Integer.MAX_VALUE;
        int sum = 0;
        for(int i=0;i<n+2;i++){
            sum += salary[i];
            max = Math.max(max,salary[i]);
            min = Math.min(min,salary[i]);
        }
        return (double)(sum-max-min)/n;
    }
}
class Solution {
public:
    double average(vector<int>& salary) {
        int n = salary.size();
        double ans=0;
        sort(salary.begin(),salary.end());
        for(int i=1;i<n-1;i++){
            ans+=salary[i];
        }
        return ans/(n-2);
    }
};
var average = function(salary) {
    let max = salary[0], min = salary[salary.length-1], sum = 0;
    for(let i = 0; i < salary.length; i++) {
        if(salary[i] > max) {
            max = salary[i];
        }
        else if(salary[i] < min) {
            min = salary[i];
        }
        sum += salary[i];
    }
    return (sum-min-max)/(salary.length-2);
};
Average Salary Excluding the Minimum and Maximum Salary