2090. K Radius Subarray Averages

Medium

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.

Example 1:

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.

Example 2:

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.

Example 3:

Input: nums = [8], k = 100000
Output: [-1]
Explanation: 
- avg[0] is -1 because there are less than k elements before and after index 0.

Constraints:

• n == nums.length

• 1 <= n <= 10^5

• 0 <= nums[i], k <= 10^5

解題

Sliding window

func getAverages(nums []int, k int) []int {
    sum := 0 // 目前 window 內總和 
    res := make([]int, len(nums))

    for i:=0; i<len(nums); i++ {
        sum += nums[i] // 加上 window 最右邊的值

        if i<k || (len(nums)-(k+1) >= 0 && i>len(nums)-(k+1)) {
            res[i] = -1
        }

        if i<2*k {
            continue
        }

        left := i - 2*k // window 的左邊
        index := left+k

        if index>=0 {
            res[index] = sum/(k*2+1)
        }
        sum -= nums[left] // 下一個 iteration window 要右移了，扣掉最左邊的值
    }

    return res
}