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Formatted question description: https://leetcode.ca/all/2465.html

# 2465. Number of Distinct Averages

• Difficulty: Easy.
• Related Topics: Array, Hash Table, Two Pointers, Sorting.
• Similar Questions: Two Sum, Finding Pairs With a Certain Sum.

## Problem

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

## Solution (Java, C++, Python)

• class Solution {
public int distinctAverages(int[] nums) {
Arrays.sort(nums);
int n = nums.length;
Set<Integer> s = new HashSet<>();
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());
int n = nums.size();
unordered_set<int> s;
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:
n = len(nums)
nums.sort()
return len(set(nums[i] + nums[n - i - 1] for i in range(n >> 1)))


• func distinctAverages(nums []int) 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)
}


Explain:

nope.

Complexity:

• Time complexity : O(n).
• Space complexity : O(n).