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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); } }
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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); } }
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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) } }