Welcome to Subscribe On Youtube

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);
    }
}

All Problems

All Solutions