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.


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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.


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Constraints:


	n == nums.length
	1 &lt;= n &lt;= 105
	0 &lt;= nums[i], k &lt;= 105


class Solution:
    def getAverages(self, nums: list[int], k: int) -> list[int]:

        n, diam = len(nums), 2*k+1
        if n < diam: return [-1]*n

        ans = [-1]*k

        arr = list(accumulate(nums, initial = 0))

        for i in range(n-diam+1):
            ans.append((arr[i+diam]-arr[i])//diam)

        return  ans + [-1]*k
class Solution
{
    public int[] getAverages(int[] nums, int k)
    {
        if(k == 0)
            return nums;

        int N = nums.length;
        long[] sum = new long[N];
        sum[0] = nums[0];

        for(int i = 1; i < N; i++)
            sum[i] = sum[i-1]+nums[i]; // Sum of 0 - ith element at sum[i]

        int[] ret = new int[N];
        Arrays.fill(ret,-1);

        for(int i = k; i < N-k; i++) // Beyond this range, there are less than k elements so -1
        {
            long temp = (sum[i+k]-sum[i-k]+nums[i-k])/(k*2+1);
            ret[i] = (int)temp;
        }
        return ret;
    }
}
class Solution {
public:
    vector<int> getAverages(vector<int>& nums, int k) {

        int n = nums.size();
        vector<int> ans(n , -1);

        if(2 * k + 1 > n) return ans;

        // Simple Sliding Window

        long long int sum = 0;

        // Take a window of size 2 * k + 1
        for(int i =0 ; i < 2 * k + 1 ; i++) {
            sum += nums[i];
        }

        ans[k] = sum / (2 * k + 1);

        // Then slide it untill the end of the window reaches at the end

        for(int i = 2 * k + 1 , j = k + 1, s = 0; i < n ; i++ , j++, s++) {

            sum += nums[i];
            sum -= nums[s];
            ans[j] = sum /(2 * k + 1);

        }

        return ans;
    }
};
var getAverages = function(nums, k) {
    const twoK = 2 * k;
    const windowSize = twoK + 1;

    const result = [...nums].fill(-1);
    let sum = 0;
    for (let i = 0; i < nums.length; i++) {
        sum += nums[i];
        if (i >= twoK) {
            result[i - k] = Math.floor(sum / windowSize)
            sum -= nums[i - twoK];
        }
    }
    return result;
};
K Radius Subarray Averages