# 2465. Number of Distinct Averages

## Description

You are given a 0-indexed integer array nums of even length.

As long as nums is not empty, you must repetitively:

• Find the minimum number in nums and remove it.
• Find the maximum number in nums and remove it.
• Calculate the average of the two removed numbers.

The average of two numbers a and b is (a + b) / 2.

• For example, the average of 2 and 3 is (2 + 3) / 2 = 2.5.

Return the number of distinct averages calculated using the above process.

Note that when there is a tie for a minimum or maximum number, any can be removed.

Example 1:

Input: nums = [4,1,4,0,3,5]
Output: 2
Explanation:
1. Remove 0 and 5, and the average is (0 + 5) / 2 = 2.5. Now, nums = [4,1,4,3].
2. Remove 1 and 4. The average is (1 + 4) / 2 = 2.5, and nums = [4,3].
3. Remove 3 and 4, and the average is (3 + 4) / 2 = 3.5.
Since there are 2 distinct numbers among 2.5, 2.5, and 3.5, we return 2.


Example 2:

Input: nums = [1,100]
Output: 1
Explanation:
There is only one average to be calculated after removing 1 and 100, so we return 1.


Constraints:

• 2 <= nums.length <= 100
• nums.length is even.
• 0 <= nums[i] <= 100

## Solutions

Solution 1: Sorting

The problem requires us to find the minimum and maximum values in the array $nums$ each time, delete them, and then calculate the average of the two deleted numbers. Therefore, we can first sort the array $nums$, then take the first and last elements of the array each time, calculate their sum, use a hash table or array $cnt$ to record the number of times each sum appears, and finally count the number of different sums.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

• class Solution {
public int distinctAverages(int[] nums) {
Arrays.sort(nums);
Set<Integer> s = new HashSet<>();
int n = nums.length;
for (int i = 0; i < n >> 1; ++i) {
s.add(nums[i] + nums[n - i - 1]);
}
return s.size();
}
}

• class Solution {
public:
int distinctAverages(vector<int>& nums) {
sort(nums.begin(), nums.end());
unordered_set<int> s;
int n = nums.size();
for (int i = 0; i < n >> 1; ++i) {
s.insert(nums[i] + nums[n - i - 1]);
}
return s.size();
}
};

• class Solution:
def distinctAverages(self, nums: List[int]) -> int:
nums.sort()
return len(set(nums[i] + nums[-i - 1] for i in range(len(nums) >> 1)))


• func distinctAverages(nums []int) (ans int) {
sort.Ints(nums)
n := len(nums)
s := map[int]struct{}{}
for i := 0; i < n>>1; i++ {
s[nums[i]+nums[n-i-1]] = struct{}{}
}
return len(s)
}

• function distinctAverages(nums: number[]): number {
nums.sort((a, b) => a - b);
const s: Set<number> = new Set();
const n = nums.length;
for (let i = 0; i < n >> 1; ++i) {
s.add(nums[i] + nums[n - i - 1]);
}
return s.size;
}


• impl Solution {
pub fn distinct_averages(nums: Vec<i32>) -> i32 {
let mut nums = nums;
nums.sort();
let n = nums.len();
let mut cnt = vec![0; 201];
let mut ans = 0;

for i in 0..n >> 1 {
let x = (nums[i] + nums[n - i - 1]) as usize;
cnt[x] += 1;

if cnt[x] == 1 {
ans += 1;
}
}

ans
}
}