Maximum Average Subarray I

Maximum Average Subarray

Naive Solution

A brute-force approach would involve iterating through all possible subarrays of length k, calculating the average of each subarray, and keeping track of the maximum average encountered. This involves nested loops, making it less efficient.

def find_max_average_naive(nums, k):
    n = len(nums)
    max_avg = float('-inf')
    for i in range(n - k + 1):
        current_sum = 0
        for j in range(i, i + k):
            current_sum += nums[j]
        current_avg = current_sum / k
        max_avg = max(max_avg, current_avg)
    return max_avg

Big O Analysis:

• Time Complexity: O(n*k), where n is the length of the input array nums. This is due to the nested loops.
• Space Complexity: O(1), as we are only using a constant amount of extra space.

Optimal Solution: Sliding Window

A more efficient approach is to use the sliding window technique. We first calculate the sum of the first k elements. Then, we slide the window by one element at a time, subtracting the element that leaves the window and adding the element that enters the window. This allows us to update the sum in O(1) time, significantly reducing the overall time complexity.

def find_max_average(nums, k):
    n = len(nums)
    current_sum = sum(nums[:k])
    max_sum = current_sum

    for i in range(k, n):
        current_sum += nums[i] - nums[i - k]
        max_sum = max(max_sum, current_sum)

    return max_sum / k

Big O Analysis:

• Time Complexity: O(n), where n is the length of the input array nums. This is because we iterate through the array only once.
• Space Complexity: O(1), as we are only using a constant amount of extra space.

Edge Cases

• Empty Array: If the input array is empty, the problem statement specifies that 1 <= k <= n, so this case is not applicable.
• k = 1: If k is 1, the maximum average is simply the maximum element in the array.
• Large Input Values: The problem statement allows values between -10^4 and 10^4. We need to consider the possibility of integer overflow if k is large and the numbers are also large. Using float can mitigate this risk, as shown in the examples above.