Question

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

Solution 1: Sorting

The problem requires us to find the minimum and maximum values in the array $nums$ each time, delete them, and then calculate the average of the two deleted numbers. Therefore, we can first sort the array $nums$, then take the first and last elements of the array each time, calculate their sum, use a hash table or array $cnt$ to record the number of times each sum appears, and finally count the number of different sums.

The time complexity is $O(n \times \log n)$, and the space complexity is $O(n)$. Here, $n$ is the length of the array $nums$.

class Solution:
  def distinctAverages(self, nums: List[int]) -> int:
    nums.sort()
    return len(set(nums[i] + nums[-i - 1] for i in range(len(nums) >> 1)))

class Solution {
  public int distinctAverages(int[] nums) {
    Arrays.sort(nums);
    Set<Integer> s = new HashSet<>();
    int n = nums.length;
    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());
    unordered_set<int> s;
    int n = nums.size();
    for (int i = 0; i < n >> 1; ++i) {
      s.insert(nums[i] + nums[n - i - 1]);
    }
    return s.size();
  }
};

func distinctAverages(nums []int) (ans 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 s: Set<number> = new Set();
  const n = nums.length;
  for (let i = 0; i < n >> 1; ++i) {
    s.add(nums[i] + nums[n - i - 1]);
  }
  return s.size;
}

impl Solution {
  pub fn distinct_averages(nums: Vec<i32>) -> i32 {
    let mut nums = nums;
    nums.sort();
    let n = nums.len();
    let mut cnt = vec![0; 201];
    let mut ans = 0;

    for i in 0..n >> 1 {
      let x = (nums[i] + nums[n - i - 1]) as usize;
      cnt[x] += 1;

      if cnt[x] == 1 {
        ans += 1;
      }
    }

    ans
  }
}

Solution 2

class Solution:
  def distinctAverages(self, nums: List[int]) -> int:
    nums.sort()
    ans = 0
    cnt = Counter()
    for i in range(len(nums) >> 1):
      x = nums[i] + nums[-i - 1]
      cnt[x] += 1
      if cnt[x] == 1:
        ans += 1
    return ans

class Solution {
  public int distinctAverages(int[] nums) {
    Arrays.sort(nums);
    int[] cnt = new int[201];
    int n = nums.length;
    int ans = 0;
    for (int i = 0; i < n >> 1; ++i) {
      if (++cnt[nums[i] + nums[n - i - 1]] == 1) {
        ++ans;
      }
    }
    return ans;
  }
}

class Solution {
public:
  int distinctAverages(vector<int>& nums) {
    sort(nums.begin(), nums.end());
    int cnt[201]{};
    int n = nums.size();
    int ans = 0;
    for (int i = 0; i < n >> 1; ++i) {
      if (++cnt[nums[i] + nums[n - i - 1]] == 1) {
        ++ans;
      }
    }
    return ans;
  }
};

func distinctAverages(nums []int) (ans int) {
  sort.Ints(nums)
  n := len(nums)
  cnt := [201]int{}
  for i := 0; i < n>>1; i++ {
    x := nums[i] + nums[n-i-1]
    cnt[x]++
    if cnt[x] == 1 {
      ans++
    }
  }
  return
}

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::HashMap;

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

    for i in 0..n >> 1 {
      let x = nums[i] + nums[n - i - 1];
      *h.entry(x).or_insert(0) += 1;

      if *h.get(&x).unwrap() == 1 {
        ans += 1;
      }
    }

    ans
  }
}

Solution 3

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
  }
}