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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)
}

  • function distinctAverages(nums: number[]): number {
    nums.sort((a, b) => a - b);
    const cnt: number[] = Array(201).fill(0);
    let ans = 0;
    const n = nums.length;
    for (let i = 0; i < n >> 1; ++i) {
        if (++cnt[nums[i] + nums[n - i - 1]] === 1) {
            ++ans;
        }
    }
    return ans;
}


  • use std::collections::HashSet;

impl Solution {
    pub fn distinct_averages(nums: Vec<i32>) -> i32 {
        let mut set = HashSet::new();
        let mut ans = 0;
        let n = nums.len();
        let mut nums = nums;
        nums.sort();

        for i in 0..n >> 1 {
            let x = nums[i] + nums[n - i - 1];

            if set.contains(&x) {
                continue;
            }

            set.insert(x);
            ans += 1;
        }

        ans
    }
}

Explain:

nope.

Complexity:

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

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