[leetcode] K Radius Subarray Averages

문제

You are given a 0-indexed array nums of n integers, and an integer k.

The k-radius average for a subarray of nums centered at some index i with the radius k is the average of all elements in nums between the indices i - k and i + k (inclusive). If there are less than k elements before or after the index i, then the k-radius average is -1.

Build and return an array avgs of length n where avgs[i] is the k-radius average for the subarray centered at index i.

The average of x elements is the sum of the x elements divided by x, using integer division. The integer division truncates toward zero, which means losing its fractional part.

• For example, the average of four elements 2, 3, 1, and 5 is (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75, which truncates to 2.

문제 예시

Input: nums = [7,4,3,9,1,8,5,2,6], k = 3
Output: [-1,-1,-1,5,4,4,-1,-1,-1]
Explanation:

• avg[0], avg[1], and avg[2] are -1 because there are less than k elements before each index.
• The sum of the subarray centered at index 3 with radius 3 is: 7 + 4 + 3 + 9 + 1 + 8 + 5 = 37.
  Using integer division, avg[3] = 37 / 7 = 5.
• For the subarray centered at index 4, avg[4] = (4 + 3 + 9 + 1 + 8 + 5 + 2) / 7 = 4.
• For the subarray centered at index 5, avg[5] = (3 + 9 + 1 + 8 + 5 + 2 + 6) / 7 = 4.
• avg[6], avg[7], and avg[8] are -1 because there are less than k elements after each index.

Input: nums = [100000], k = 0
Output: [100000]
Explanation:

• The sum of the subarray centered at index 0 with radius 0 is: 100000.
  avg[0] = 100000 / 1 = 100000.

Input: nums = [8], k = 100000
Output: [-1]
Explanation:

• avg[0] is -1 because there are less than k elements before and after index 0.

제한사항

• n == nums.length
• 1 <= n <= 105
• 0 <= nums[i], k <= 105

풀이 접근 방식

시간 및 공간 복잡도

소스코드

class Solution:
    def getAverages(self, nums: List[int], k: int) -> List[int]:
        
        n = len(nums)
        prefix = [nums[0]]
        for i in range(1, n):
            prefix.append(nums[i] + prefix[-1])
        
        ans = [-1] * n
        window_size = 2 * k + 1
        for i in range(k, n-k):
            ans[i] = (prefix[i+k] - prefix[i-k] + nums[i-k]) // window_size
        
        return ans