LeetCode 2090. K Radius Subarray Averages Solution in Java, C++, Python & More | Explanation + Code

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:

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

Solutions

Solution 1: Sliding Window

The length of a subarray with radius k is k × 2 + 1, so we can maintain a window of size k × 2 + 1 and denote the sum of all elements in the window as s.

We create an answer array ans of length n, initially setting each element to -1.

Next, we traverse the array nums, adding the value of nums[i] to the window sum s. If i ≥ k × 2, it means the window size is k × 2 + 1, so we set ans[i-k] = ^s⁄_{k × 2 + 1}. Then, we remove the value of nums[i - k × 2] from the window sum s. Continue traversing the next element.

Finally, return the answer array.

The time complexity is O(n), where n is the length of the array nums. Ignoring the space consumption of the answer array, the space complexity is O(1).

PythonJavaC++GoTypeScript

class Solution: def getAverages(self, nums: List[int], k: int) -> List[int]: n = len(nums) ans = [-1] * n s = 0 for i, x in enumerate(nums): s += x if i >= k * 2: ans[i - k] = s // (k * 2 + 1) s -= nums[i - k * 2] return ans(code-box)

class Solution { public int[] getAverages(int[] nums, int k) { int n = nums.length; int[] ans = new int[n]; Arrays.fill(ans, -1); long s = 0; for (int i = 0; i < n; ++i) { s += nums[i]; if (i >= k * 2) { ans[i - k] = (int) (s / (k * 2 + 1)); s -= nums[i - k * 2]; } } return ans; } }(code-box)

class Solution { public: vector<int> getAverages(vector<int>& nums, int k) { int n = nums.size(); vector<int> ans(n, -1); long long s = 0; for (int i = 0; i < n; ++i) { s += nums[i]; if (i >= k * 2) { ans[i - k] = s / (k * 2 + 1); s -= nums[i - k * 2]; } } return ans; } };(code-box)

func getAverages(nums []int, k int) []int { ans := make([]int, len(nums)) for i := range ans { ans[i] = -1 } s := 0 for i, x := range nums { s += x if i >= k*2 { ans[i-k] = s / (k*2 + 1) s -= nums[i-k*2] } } return ans }(code-box)

function getAverages(nums: number[], k: number): number[] { const n = nums.length; const ans: number[] = Array(n).fill(-1); let s = 0; for (let i = 0; i < n; ++i) { s += nums[i]; if (i >= k * 2) { ans[i - k] = Math.floor(s / (k * 2 + 1)); s -= nums[i - k * 2]; } } return ans; }(code-box)