[LeetCode] 2090. K Radius Subarray Averages

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 <= 105
0 <= nums[i], k <= 105

半径为 k 的子数组平均值。

给你一个下标从 0 开始的数组 nums ，数组中有 n 个整数，另给你一个整数 k 。

半径为 k 的子数组平均值 是指：nums 中一个以下标 i 为 中心 且 半径 为 k 的子数组中所有元素的平均值，即下标在 i - k 和 i + k 范围（含 i - k 和 i + k）内所有元素的平均值。如果在下标 i 前或后不足 k 个元素，那么 半径为 k 的子数组平均值 是 -1 。

构建并返回一个长度为 n 的数组 avgs ，其中 avgs[i] 是以下标 i 为中心的子数组的 半径为 k 的子数组平均值 。

x 个元素的 平均值 是 x 个元素相加之和除以 x ，此时使用截断式 整数除法 ，即需要去掉结果的小数部分。

例如，四个元素 2、3、1 和 5 的平均值是 (2 + 3 + 1 + 5) / 4 = 11 / 4 = 2.75，截断后得到 2 。

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/k-radius-subarray-averages
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思路

思路是前缀和。求一段子数组的和的做法是前缀和，这应该成为刷题的本能反应。

我们首先用一个额外数组记录整个数组的前缀和，然后再次遍历数组，根据条件计算以每一个下标 i 为中心，半径为 k 的子数组的和。

复杂度

时间O(n)
空间O(n)

代码

Java实现

class Solution {
    public int[] getAverages(int[] nums, int k) {
        int n = nums.length;
        // 要用long型，不然超出范围
        long[] presum = new long[n + 1];
        for (int i = 0; i < n; i++) {
            presum[i + 1] = presum[i] + nums[i];
        }

        int[] res = new int[n];
        Arrays.fill(res, -1);
        for (int i = 0; i < n; i++) {
            if (i - k >= 0 && i + k < n) {
                res[i] = (int)((presum[i + k + 1] - presum[i - k]) / (2 * k + 1));
            }
        }
        return res;
    }
}