How do you solve Maximum Average Subarray I on LeetCode?

Maximum Average Subarray I (LeetCode #643) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(1) space complexity. Key insight: Using a sliding window can significantly reduce the time complexity from O(n²) to O(n).

What is the brute force solution for Maximum Average Subarray I?

The brute force approach for Maximum Average Subarray I is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves checking every possible subarray of length k and calculating their averages. This 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 Maximum Average Subarray I?

Maximum Average Subarray I uses the following concepts: Array, Sliding Window. The recommended patterns to study are: Sliding Window, Array.

What are the interview tips for Maximum Average Subarray I?

Always clarify the constraints and edge cases before diving into coding. Explain your thought process and the trade-offs of different approaches. Practice writing clean and efficient code, as readability is crucial in interviews.

What are common mistakes when solving Maximum Average Subarray I?

Not considering edge cases where k equals 1 or n. Failing to update the sum correctly while sliding the window.

#643

Maximum Average Subarray I

Easy

ArraySliding WindowSliding WindowArray

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

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

💡

Intuition

Time O(n)Space O(1)

The optimal approach uses a sliding window technique to maintain the sum of the current subarray of length k. This allows us to compute the maximum average in linear time by reusing the previous sum.

⚙️

Algorithm

4 steps

1Step 1: Calculate the sum of the first k elements.
2Step 2: Initialize max_avg with the average of the first k elements.
3Step 3: Slide the window across the array by adding the next element and removing the first element of the previous window.
4Step 4: Update max_avg whenever a new average is greater than the current max_avg.

solution.py9 lines

1# Full working Python code
2
3def findMaxAverage(nums, k):
4    current_sum = sum(nums[:k])
5    max_avg = current_sum / k
6    for i in range(k, len(nums)):
7        current_sum += nums[i] - nums[i - k]
8        max_avg = max(max_avg, current_sum / k)
9    return max_avg

ℹ

Complexity note: This complexity is O(n) because we only traverse the array once, maintaining a running sum without needing additional space for storing subarrays.

1Using a sliding window can significantly reduce the time complexity from O(n²) to O(n).
2Maintaining a running sum allows for efficient updates as the window slides.

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