LeetCode-in-Java

2465. Number of Distinct Averages

Easy

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

import java.util.Arrays;
import java.util.HashSet;
import java.util.Set;

public class Solution {
    public int distinctAverages(int[] nums) {
        Arrays.sort(nums);
        Set<Integer> set = new HashSet<>();
        int l = 0;
        int r = nums.length - 1;
        while (l < r) {
            set.add(nums[l++] + nums[r--]);
        }
        return set.size();
    }
}

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