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

Solution 1

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

Java Code
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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);
    }
}

C++ Code
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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);
    }
};

Go Code
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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)
}

TypeScript Code
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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);
}

Rust Code
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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)
    }
}

PHP Code
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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);
    }
}

C Code
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#define max(a, b) (((a) > (b)) ? (a) : (b))
#define min(a, b) (((a) < (b)) ? (a) : (b))

double average(int* salary, int salarySize) {
    int ma = INT_MIN;
    int mi = INT_MAX;
    int sum = 0;
    for (int i = 0; i < salarySize; i++) {
        sum += salary[i];
        ma = max(ma, salary[i]);
        mi = min(mi, salary[i]);
    }
    return (sum - mi - ma) * 1.0 / (salarySize - 2);
}