# 1491. Average Salary Excluding the Minimum and Maximum Salary

## Description

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.

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


Constraints:

• 3 <= salary.length <= 100
• 1000 <= salary[i] <= 106
• All the integers of salary are unique.

## Solutions

• class Solution {
public double average(int[] salary) {
int s = 0;
int mi = 10000000, mx = 0;
for (int v : salary) {
mi = Math.min(mi, v);
mx = Math.max(mx, v);
s += v;
}
s -= (mi + mx);
return s * 1.0 / (salary.length - 2);
}
}

• class Solution {
public:
double average(vector<int>& salary) {
int s = 0;
int mi = 1e7, mx = 0;
for (int v : salary) {
s += v;
mi = min(mi, v);
mx = max(mx, v);
}
s -= (mi + mx);
return (double) s / (salary.size() - 2);
}
};

• class Solution:
def average(self, salary: List[int]) -> float:
s = sum(salary) - min(salary) - max(salary)
return s / (len(salary) - 2)


• func average(salary []int) float64 {
s := 0
mi, mx := 10000000, 0
for _, v := range salary {
s += v
mi = min(mi, v)
mx = max(mx, v)
}
s -= (mi + mx)
return float64(s) / float64(len(salary)-2)
}

• function average(salary: number[]): number {
let max = -Infinity;
let min = Infinity;
let sum = 0;
for (const v of salary) {
sum += v;
max = Math.max(max, v);
min = Math.min(min, v);
}
return (sum - max - min) / (salary.length - 2);
}


• class Solution {
/**
* @param Integer[] $salary * @return Float */ function average($salary) {
$max =$sum = 0;
$min = 10 ** 6; for ($i = 0; $i < count($salary); $i++) {$min = min($min,$salary[$i]);$max = max($max,$salary[$i]);$sum += $salary[$i];
}
return ($sum -$max - $min) / (count($salary) - 2);
}
}


• impl Solution {
pub fn average(salary: Vec<i32>) -> f64 {
let n = salary.len() as i32;
let mut min = i32::MAX;
let mut max = i32::MIN;
let mut sum = 0;
for &num in salary.iter() {
min = min.min(num);
max = max.max(num);
sum += num;
}
f64::from(sum - min - max) / f64::from(n - 2)
}
}