2090. K Radius Subarray Averages

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2090. K Radius Subarray Averages

Description

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:

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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:

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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:

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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

Hints/Notes

  • 2024/10/05
  • sliding window
  • No solution from 0x3F

Solution

Language: C++

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class Solution {
public:
    vector<int> getAverages(vector<int>& nums, int k) {
        int n = nums.size(), right = 0, len = 2 * k + 1;
        long sum = 0;
        vector<int> res(n, -1);
        while (right < n) {
            sum += nums[right];
            if (right >= len - 1) {
                res[right - k] = sum / len;
                sum -= nums[right - len + 1];
            }
            right++;
        }
        return res;
    }
};