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This approach involves iterating over each element in the array and calculating the subarray sum for each index considering the radius. If there is an insufficient number of elements, return -1 for that index. The average is computed using integer division.
Time Complexity: O(n*k), where n is the size of the array.
Space Complexity: O(n), for the output array.
1#include <vector>
2#include <iostream>
3
4std::vector<int> getAverages(std::vector<int>& nums, int k) {
5 int n = nums.size();
6 std::vector<int> avgs(n, -1);
7 int subArraySize = 2 * k + 1;
8
9 for (int i = k; i < n - k; ++i) {
10 long sum = 0;
11 for (int j = i - k; j <= i + k; ++j) {
12 sum += nums[j];
13 }
14 avgs[i] = sum / subArraySize;
15 }
16
17 return avgs;
18}
19
20int main() {
21 std::vector<int> nums = {7, 4, 3, 9, 1, 8, 5, 2, 6};
22 int k = 3;
23 std::vector<int> averages = getAverages(nums, k);
24
25 for (auto avg : averages) {
26 std::cout << avg << " ";
27 }
28 return 0;
29}This C++ solution also implements a brute force method where the sum for each index's subarray is calculated directly, excluding indices where a complete k-radius subarray isn't possible.
The sliding window approach helps optimize the brute force method by avoiding redundant calculations when sums overlap, significantly reducing the time complexity.
Time Complexity: O(n), as each element is added and removed from the sum at most once.
Space Complexity: O(n), for storing the results.
1import
This Java implementation effectively uses sliding window optimization to keep a running total sum for the needed window range, achieving linear time complexity.