2090. K Radius Subarray Averages

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# 題目敘述

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.

# 解題思路

# Solution

import java.util.Arrays;

class Solution {

    public int[] getAverages(int[] nums, int k) {

        int n = nums.length;

        int windows = 2 * k + 1;

        int[] ans = new int[n];

        Arrays.fill(ans, -1);

        if(n < windows){

            return ans;

        }

        long[] sum = new long[n + 1];

        for(int i = 0; i < n; i++){

            sum[i + 1] = sum[i] + nums[i];

        } 

        for(int i = k; i < n - k; i++){

            ans[i] = (int) ((sum[i + k + 1] - sum[i - k]) / windows);

        }

        return ans;

    }

}