2465. Number of Distinct Averages

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)

class Solution {
    public int distinctAverages(int[] nums) {
        HashMap<Float,Integer> out= new  HashMap<Float,Integer>();
        Arrays.sort(nums);
        for(int i=0;i<nums.length/2;i++)
        {
            out.put((float)(nums[i]+nums[nums.length-i-1])/2,1);
        }

        return out.size();
    }
}

Explain:

nope.

Complexity:

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