You are given a 0-indexed array nums of n integers, and an integer k.
The k-radius average for a subarray of numscentered at some index i with the radiusk 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 arrayavgsof lengthnwhereavgs[i]is the k-radius average for the subarray centered at indexi.
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.
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.
To efficiently compute the average of every subarray of length 2k+1 centered at each index, we use a prefix sum array. This allows us to get the sum of any subarray in constant time. If there are not enough elements before or after an index, the answer is -1 for that index.
classSolution {
public: vector<int> getAverages(vector<int>& nums, int k) {
int n = nums.size();
vector<int> ans(n, -1);
if (2* k +1> n) return ans;
vector<longlong> prefix(n +1, 0);
for (int i =0; i < n; ++i) prefix[i +1] = prefix[i] + nums[i];
for (int i = k; i + k < n; ++i) {
longlong sum = prefix[i + k +1] - prefix[i - k];
ans[i] = sum / (2* k +1);
}
return ans;
}
};
classSolution {
publicint[]getAverages(int[] nums, int k) {
int n = nums.length;
int[] ans =newint[n];
Arrays.fill(ans, -1);
if (2 * k + 1 > n) return ans;
long[] prefix =newlong[n + 1];
for (int i = 0; i < n; i++) prefix[i + 1]= prefix[i]+ nums[i];
for (int i = k; i + k < n; i++) {
long sum = prefix[i + k + 1]- prefix[i - k];
ans[i]= (int)(sum / (2 * k + 1));
}
return ans;
}
}
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classSolution {
fungetAverages(nums: IntArray, k: Int): IntArray {
val n = nums.size
val ans = IntArray(n) { -1 }
if (2 * k + 1 > n) return ans
val prefix = LongArray(n + 1)
for (i in0 until n) prefix[i + 1] = prefix[i] + nums[i]
for (i in k until n - k) {
val sum = prefix[i + k + 1] - prefix[i - k]
ans[i] = (sum / (2 * k + 1)).toInt()
}
return ans
}
}
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classSolution:
defgetAverages(self, nums: list[int], k: int) -> list[int]:
n = len(nums)
ans = [-1] * n
if2* k +1> n:
return ans
prefix = [0]
for num in nums:
prefix.append(prefix[-1] + num)
for i in range(k, n - k):
s = prefix[i + k +1] - prefix[i - k]
ans[i] = s // (2* k +1)
return ans
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impl Solution {
pubfnget_averages(nums: Vec<i32>, k: i32) -> Vec<i32> {
let n = nums.len();
letmut ans =vec![-1; n];
let k = k asusize;
if2* k +1> n {
return ans;
}
letmut prefix =vec![0i64; n +1];
for i in0..n {
prefix[i +1] = prefix[i] + nums[i] asi64;
}
for i in k..n - k {
let sum = prefix[i + k +1] - prefix[i - k];
ans[i] = (sum / (2* k +1) asi64) asi32;
}
ans
}
}