2465. Number of Distinct Averages

문제)

2465. Number of Distinct Averages

 

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

 

솔루션1)

- 오름차순 정렬해서 제일 왼쪽은 min val, 제일 오른쪽은 max val을 쉽게 구할 수 있음

class Solution:
    def distinctAverages(self, nums: List[int]) -> int:        
        nums.sort()
        avgs = []
        while nums:
            avg = (nums[0] + nums[-1]) / 2
            avgs.append(avg)
            del nums[0]            
            nums.pop()
        
        return len(set(avgs))

솔루션2)

- 솔루션1 최적화

  : 리스트의 원소를 삭제하지 않고 left, right 인덱스 이동

  : 불필요한 나누기 2 연산 제거

class Solution:
    def distinctAverages(self, nums: List[int]) -> int:                
        nums.sort()
        l, r = 0, len(nums) - 1
        
        averages = []
        while l < r:
            averages.append(nums[l] + nums[r])
            l += 1
            r -= 1
        return len(set(averages))