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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.
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); } }
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// 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); } };
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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)
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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) } 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 }
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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); }
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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) } }
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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); } }