How do you solve K Radius Subarray Averages on LeetCode?

K Radius Subarray Averages (LeetCode #2090) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Using a sliding window can significantly reduce time complexity.

What is the brute force solution for K Radius Subarray Averages?

The brute force approach for K Radius Subarray Averages is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach calculates the k-radius average by iterating through each index and summing the elements in the range from i-k to i+k. This method is straightforward but inefficient for larger arrays. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve K Radius Subarray Averages?

K Radius Subarray Averages uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Prefix Sums.

What are the interview tips for K Radius Subarray Averages?

Always clarify edge cases with the interviewer. Explain your thought process while coding to demonstrate understanding. Practice writing clean and efficient code.

What are common mistakes when solving K Radius Subarray Averages?

Not checking boundary conditions for the indices properly. Forgetting to handle integer division correctly.

#2090

K Radius Subarray Averages

Medium

ArraySliding WindowSliding WindowPrefix Sums

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

💡

Intuition

Time O(n)Space O(n)

The optimal solution uses a sliding window technique to maintain the sum of the current k-radius subarray, allowing us to compute averages in constant time after the initial sum. This significantly reduces the time complexity.

⚙️

Algorithm

3 steps

1Step 1: Initialize an empty result array avgs of the same length as nums and a variable total to keep track of the current sum.
2Step 2: Calculate the sum of the first 2k+1 elements (if possible) and store the average for the center index k.
3Step 3: Slide the window one element at a time, updating the total by subtracting the element that is sliding out and adding the new element that is sliding in, and calculate the new average.

solution.py11 lines

1def getAverages(nums, k):
2    n = len(nums)
3    avgs = [-1] * n
4    if 2 * k + 1 > n:
5        return avgs
6    total = sum(nums[:2 * k + 1])
7    avgs[k] = total // (2 * k + 1)
8    for i in range(k + 1, n - k):
9        total += nums[i + k] - nums[i - k - 1]
10        avgs[i] = total // (2 * k + 1)
11    return avgs

ℹ

Complexity note: The time complexity is O(n) because we traverse the array once to calculate the initial sum and then slide the window across the array. The space complexity is O(n) due to the output array avgs.

1Using a sliding window can significantly reduce time complexity.
2Understanding when to apply prefix sums or sliding window techniques is crucial.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.