1491 - Average Salary Excluding the Minimum and Maximum Salary (Easy)

Problem Link

https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/

Problem Statement

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] <= 10^6
• All the integers of salary are unique.

Approach 1: One Pass

We iterate $salary$ and calculate the total sum $sum$, the minimum value $mi$ and the maximum value $mx$. The average salary of employees excluding the minimum and maximum salary would be $sum - mi - mx$. Since we exclude two items, we just need to divide by $len(salary) - 2$.

Written by @wingkwong

func average(salary []int) float64 {
    sum, n := 0, len(salary)
    mi, mx := 100005, 0
    for _, x := range salary {
        // calculate the total sum
        sum += x
        // find the minimum value
        if x < mi { mi = x }
        // find the maximum value
        if x > mx { mx = x }
    }
    // exclude mi & mx and get the avg 
    return float64(sum - mi - mx) / float64(n - 2)
}

Written by @wingkwong

use std::cmp;

impl Solution {
    pub fn average(salary: Vec<i32>) -> f64 {
        let mut sum = 0;
        let mut mx = 0;
        let mut mi = 100005;
        let n = salary.len() as f64;
        for x in salary {
            sum += x;
            mx = cmp::max(mx, x);
            mi = cmp::min(mi, x);
        }
        return ((sum - mx - mi) as f64 / (n - 2.0)); 
    }
}