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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.
1def getAverages(nums, k):
2 n = len(nums)
3 avgs = [-1] * n
4 sub_array_size = 2 * k + 1
5
6 for i in range(k, n - k):
7 total = 0
8 for j in range(i - k, i + k + 1):
9 total += nums[j]
10 avgs[i] = total // sub_array_size
11
12 return avgs
13
14nums = [7, 4, 3, 9, 1, 8, 5, 2, 6]
15k = 3
16print(getAverages(nums, k))This Python solution follows a brute force pattern, using nested loops to sum the elements within the k-radius around each valid index, and sets -1 where not applicable.
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.
1#include <vector>
#include <iostream>
std::vector<int> getAverages(std::vector<int>& nums, int k) {
int n = nums.size();
std::vector<int> avgs(n, -1);
int subArraySize = 2 * k + 1;
long long windowSum = 0;
for (int i = 0; i < n; ++i) {
windowSum += nums[i];
if (i >= subArraySize) {
windowSum -= nums[i - subArraySize];
}
if (i >= subArraySize - 1) {
avgs[i - k] = windowSum / subArraySize;
}
}
return avgs;
}
int main() {
std::vector<int> nums = {7, 4, 3, 9, 1, 8, 5, 2, 6};
int k = 3;
std::vector<int> averages = getAverages(nums, k);
for (auto avg : averages) {
std::cout << avg << " ";
}
return 0;
}The C++ implementation uses a sliding window technique, maintaining a running sum by adding the new element and subtracting the old one beyond the window size.