2090.K Radius Subarray Averages

tags: Medium,Array,Sliding Window

2090. K Radius Subarray Averages

題目描述

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.

範例

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

解答

Javascript

function getAverages(nums, k) {
  const result = Array(nums.length).fill(-1);
  const length = 2 * k + 1;
  let sum = 0;
  let left = 0;

  for (let i = 0; i < nums.length; i++) {
    sum += nums[i];
    if (i >= length - 1) {
      result[i - k] = Math.floor(sum / length);
      sum -= nums[left];
      left++;
    }
  }

  return result;
}

記錄一下第一次使用leetcode的計時功能，從讀題到AC共花15分3秒。
MarsgoatTue, Jun 20, 2023

C++

using LL = long long int;
class Solution {
public:
    vector<int> getAverages(vector<int>& nums, int k) {
        ios_base::sync_with_stdio(0); cin.tie(0);
        if (k == 0) {
            return nums;
        }
        int n = nums.size();
        vector<LL> prefixSum(n + 1, 0);
        prefixSum[1] = nums[0];
        for (int i = 2; i < n + 1; i ++) {
            prefixSum[i] = prefixSum[i - 1] + nums[i - 1];
        }
        vector<int> ans(n, -1);
        for (int i = 0; i < n; i ++) {
            if (i - k < 0 or i + k >= n) {
                continue;
            }
            ans[i] = (prefixSum[i + k + 1] - prefixSum[i - k]) / (2 * k + 1);
        }
        return ans;
    }
};

index	0	1	2	3	4	5	6	7	8	9
nums	7	4	3	9	1	8	5	2	6
prefixSum	0	7	11	14	23	24	32	37	39	45

Jerry Wu20 June, 2023

C#

public class Solution {
    public int[] GetAverages(int[] nums, int k) {
        if (k == 0) return nums;     
        int n = nums.Length;
        int slide = k * 2 + 1;
        
        int[] ans = new int[n];
        long sum = 0;
        for (int i = 0; i < n; i++) {
            ans[i] = -1;
            sum += nums[i];
            if (i >= slide - 1) {
                ans[i - k] = (int)(sum / slide);
                sum -= nums[i - slide + 1];
            }
        }
        return ans;
    }
}

JimJun 20, 2023

Reference

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