#2090

Medium

auto_awesomeSliding window with running state updates

LeetCode Problem Workspace

K Radius Subarray Averages

Efficiently calculate the k-radius average for subarrays using sliding window technique.

Problem Statement

You are given an array of integers nums and an integer k. For each index i in the array, compute the k-radius average, which is the average of elements between indices i - k and i + k, inclusive. If there are fewer than k elements before or after index i, the result is -1.

Return an array where the value at each index represents the k-radius average for the subarray centered at that index. If the subarray cannot be formed due to insufficient elements, return -1 for those indices.

Examples

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]

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]

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]

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

Solution Approach

Sliding Window for Running Sum

To calculate the average of a subarray efficiently, maintain a running sum of the elements in the current window. Move the window across the array while adding the new element at the end and subtracting the element at the start of the window to maintain an accurate sum. This approach avoids recalculating the sum from scratch for each subarray, leading to a time complexity of O(n).

Handling Edge Cases with Boundaries

For each index i, ensure that the window is within bounds. If there are fewer than k elements before or after index i, return -1 for those indices. This involves checking the index ranges and adjusting the window accordingly, which ensures that the solution handles all edge cases efficiently.

Optimized Time Complexity

By using the sliding window technique, we can calculate the k-radius average for all indices in linear time. Instead of recalculating the sum for each subarray from scratch, the window dynamically updates, maintaining an O(n) time complexity. This optimization is crucial for large inputs where brute force approaches would be too slow.

Complexity Analysis

Metric	Value
Time	Depends on the final approach
Space	Depends on the final approach

The time complexity of the solution is O(n) because we process each element once as we slide the window across the array. The space complexity is O(1) as we only maintain a few variables for the window sum and the result array, regardless of the input size.

What Interviewers Usually Probe

Test if the candidate can efficiently manage window updates for large inputs.
Evaluate if the candidate handles boundary conditions (less than k elements before or after index).
Assess the candidate's understanding of sliding window techniques and optimization.

Common Pitfalls or Variants

Common pitfalls

Failing to correctly handle the boundary conditions when there are fewer than k elements before or after an index.
Not maintaining an efficient running sum in the sliding window, leading to unnecessary recomputation of sums.
Overlooking the time complexity and attempting a brute force solution that recalculates the sum for each subarray.

Follow-up variants

Generalize the solution to handle variable radius sizes.
Consider a variant where the input array contains negative numbers, testing if the candidate adapts the sliding window technique.
Modify the problem to compute a different statistic, such as the product or median, within the window.

FAQ

What is the time complexity of the K Radius Subarray Averages problem?

The time complexity is O(n) because we use a sliding window technique that processes each element exactly once.

How do you handle cases where there are fewer than k elements before or after an index?

In these cases, the k-radius average for that index is set to -1, as there are insufficient elements to form the subarray.

Can this problem be solved with a brute force approach?

While a brute force approach is possible, it would have a time complexity of O(n*k), which is inefficient for large arrays. Sliding window optimization reduces this to O(n).

What are the constraints for the K Radius Subarray Averages problem?

The array length n can be as large as 10^5, and the values of k and nums[i] range from 0 to 10^5.

How do I calculate the sum for a subarray using sliding window?

By maintaining a running sum of the current window, you add the new element entering the window and subtract the old element leaving the window as you slide across the array.

terminal

Solution

Solution 1: Sliding Window

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

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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

Continue Practicing

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