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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.

Related Topics:
Array, Sort

## Solution 1.

• 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);
}
}

############

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);
}
}

• // 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)


• 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)
}

func max(a, b int) int {
if a > b {
return a
}
return b
}

func min(a, b int) int {
if a < b {
return a
}
return b
}

• 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);
}


• 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)
}
}


• 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);
}
}