README_EN.md

2465. Number of Distinct Averages

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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

Python3

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)))

Java

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();
    }
}

C++

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();
    }
};

Go

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)
}

TypeScript

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