80 lines
1.8 KiB
Markdown
80 lines
1.8 KiB
Markdown
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# Leetcode Average-Salary-Excluding-the-Minimum-and-Maximum-Salary
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#### 2022-07-23 15:27
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> ##### Difficulty:
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>
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> #coding_problems #difficulty_easy
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>
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> ##### Additional tags:
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>
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> #leetcode #math
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>
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> ##### Revisions:
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>
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> N/A
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##### Related topics:
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##### Links:
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- [Link to problem](https://leetcode.com/problems/average-salary-excluding-the-minimum-and-maximum-salary/)
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***
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### Problem
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You are given an array of **unique** integers `salary` where `salary[i]` is the salary of the `ith` employee.
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Return _the average salary of employees excluding the minimum and maximum salary_. Answers within `10-5` of the actual answer will be accepted.
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#### Examples
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**Example 1:**
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**Input:** salary = [4000,3000,1000,2000]
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**Output:** 2500.00000
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**Explanation:** Minimum salary and maximum salary are 1000 and 4000 respectively.
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Average salary excluding minimum and maximum salary is (2000+3000) / 2 = 2500
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**Example 2:**
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**Input:** salary = [1000,2000,3000]
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**Output:** 2000.00000
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**Explanation:** Minimum salary and maximum salary are 1000 and 3000 respectively.
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Average salary excluding minimum and maximum salary is (2000) / 1 = 2000
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#### Constraints
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- `3 <= salary.length <= 100`
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- `1000 <= salary[i] <= 106`
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- All the integers of `salary` are **unique**.
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### Thoughts
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> [!summary]
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> This is a #math problem
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The one solution I came up with is using math.
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When counting total, be warned of integer overflows.
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### Solution
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```cpp
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class Solution {
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public:
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double average(vector<int> &salary) {
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double ans = 0, factor = salary.size() - 2;
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int maxSal = 0, minSal = 1000001;
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for (int i = 0, size = factor + 2; i < size; i++) {
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ans += salary[i] / factor;
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maxSal = max(maxSal, salary[i]);
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minSal = min(minSal, salary[i]);
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}
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return ans - minSal / factor - maxSal / factor;
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}
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};
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```
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