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2465. Number of Distinct Averages

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 $nums$.

• Java
• C++
• Python
• Go
• TypeScript
• RenderScript

• 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();
      }
  };
  

• 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)))
  
  

• 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
      }
  }
  
  

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