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Maximum Average Subarray I

Easy

Maximum Average Subarray I

Easy

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type of the numbers in the input array and the value of 'k'? Can they be floating-point numbers or only integers?
Can 'k' be larger than the size of the input array, or less than or equal to 0?
Is the input array guaranteed to have at least 'k' elements, or should I handle the case where it has fewer?
What should I return if the input array is empty or null?
Are there any specific requirements for the precision of the average that I need to consider for floating-point numbers?

Brute Force Solution

Approach

We need to find the best small group of numbers within a larger list of numbers. The brute force approach looks at every single possible group of the required size, without skipping any.

Here's how the algorithm would work step-by-step:

First, consider the first small group of numbers from the beginning of the list.
Calculate the average value of this group.
Next, shift one number to the right, forming a new group, and calculate its average value.
Continue sliding this group one number at a time across the entire list, calculating the average each time.
Keep track of the highest average value you've seen so far.
After considering every possible group, the highest average you've tracked is the answer.

Code Implementation

def find_max_average_brute_force(number_list, subarray_length):
    maximum_average = float('-inf')

    # Iterate through all possible starting positions of the subarray
    for i in range(len(number_list) - subarray_length + 1):

        # Calculate the sum of the current subarray
        current_subarray_sum = 0
        for j in range(subarray_length):
            current_subarray_sum += number_list[i + j]

        #Calculate average of current subarray
        current_average = current_subarray_sum / subarray_length

        # Update maximum_average if current average is greater
        maximum_average = max(maximum_average, current_average)

    return maximum_average

Big(O) Analysis

Time Complexity

O(n*k) – The algorithm iterates through the array of size n, and for each starting position, it calculates the average of a subarray of size k. Calculating the average of each subarray of size k requires iterating through k elements. Therefore, the dominant operation is iterating k times for each of the n possible starting positions. The total number of operations is proportional to n * k, resulting in a time complexity of O(n*k).

Space Complexity

O(1) – The provided plain English explanation calculates averages by sliding a window across the input list and keeping track of the maximum average seen so far. It doesn't describe the creation of any auxiliary data structures such as lists, hash maps, or trees. It only mentions calculating and storing a single maximum average value. Thus the algorithm uses a fixed amount of extra memory irrespective of the input size N (where N is the length of the list), leading to a constant space complexity.

Optimal Solution

Approach

The most efficient way to find the maximum average of a continuous group of numbers is by using a sliding window technique. Imagine a window of a fixed size moving across the numbers. By only updating the sum within the window, we avoid recalculating the entire average each time, making the process much faster.

Here's how the algorithm would work step-by-step:

First, calculate the sum of the initial group of numbers that matches the required size.
Then, move the window one position to the right by subtracting the leftmost number from the previous group and adding the new rightmost number to the group's sum. This avoids recalculating the whole sum.
While moving the window, keep track of the highest sum you've seen so far.
After examining every group of numbers, divide the highest sum by the size of the group to get the maximum average.

Code Implementation

def find_max_average(numbers, subarray_length):
    current_subarray_sum = sum(numbers[:subarray_length])
    max_subarray_sum = current_subarray_sum

    # Start sliding window after initial subarray
    for i in range(subarray_length, len(numbers)):
        current_subarray_sum += numbers[i] - numbers[i - subarray_length]

        # Keep track of the largest subarray sum
        max_subarray_sum = max(max_subarray_sum, current_subarray_sum)

    # Calculating average after iterating saves time
    maximum_average = max_subarray_sum / subarray_length
    return maximum_average

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums of size n once. The initial sum calculation takes O(k) time where k is the size of the window, but k is a constant. The sliding window then moves across the array performing a constant number of operations (subtraction and addition) for each position. Therefore, the time complexity is dominated by the single pass through the array, resulting in O(n).

Space Complexity

O(1) – The provided solution uses a sliding window approach. It calculates a running sum and keeps track of the maximum sum encountered so far. These operations use a fixed number of variables (for the current sum and maximum sum), irrespective of the input array's size, N. Therefore, the algorithm utilizes constant auxiliary space, resulting in O(1) space complexity.

Edge Cases

Case	How to Handle
Empty input array	Return 0 or throw an IllegalArgumentException as there's no subarray.
k is larger than the array size	Return 0 or throw an IllegalArgumentException as a subarray of size k is impossible.
Array size equals k (subarray is the whole array)	Calculate the average of the entire array.
Array contains large positive and negative numbers leading to potential overflow when summing	Use long type to store the sum to avoid integer overflow.
k is 1 (single element subarray)	Return the maximum element in the array, as it will have the highest average.
All elements in the array are negative	The algorithm should correctly identify the subarray with the least negative (closest to zero) average.
All elements in the array are zero	The average will be zero regardless of the subarray, so return 0.
Floating-point precision issues with very large or small average values	Consider using double for intermediate calculations and result to minimize precision loss.

Solution

Clarifying Questions

When you get asked this question in a real-life environment, it will often be ambiguous (especially at FAANG). Make sure to ask these questions in that case:

What is the expected data type of the numbers in the input array and the value of 'k'? Can they be floating-point numbers or only integers?
Can 'k' be larger than the size of the input array, or less than or equal to 0?
Is the input array guaranteed to have at least 'k' elements, or should I handle the case where it has fewer?
What should I return if the input array is empty or null?
Are there any specific requirements for the precision of the average that I need to consider for floating-point numbers?

Brute Force Solution

Approach

We need to find the best small group of numbers within a larger list of numbers. The brute force approach looks at every single possible group of the required size, without skipping any.

Here's how the algorithm would work step-by-step:

First, consider the first small group of numbers from the beginning of the list.
Calculate the average value of this group.
Next, shift one number to the right, forming a new group, and calculate its average value.
Continue sliding this group one number at a time across the entire list, calculating the average each time.
Keep track of the highest average value you've seen so far.
After considering every possible group, the highest average you've tracked is the answer.

Code Implementation

def find_max_average_brute_force(number_list, subarray_length):
    maximum_average = float('-inf')

    # Iterate through all possible starting positions of the subarray
    for i in range(len(number_list) - subarray_length + 1):

        # Calculate the sum of the current subarray
        current_subarray_sum = 0
        for j in range(subarray_length):
            current_subarray_sum += number_list[i + j]

        #Calculate average of current subarray
        current_average = current_subarray_sum / subarray_length

        # Update maximum_average if current average is greater
        maximum_average = max(maximum_average, current_average)

    return maximum_average

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

Here's how the algorithm would work step-by-step:

First, calculate the sum of the initial group of numbers that matches the required size.
Then, move the window one position to the right by subtracting the leftmost number from the previous group and adding the new rightmost number to the group's sum. This avoids recalculating the whole sum.
While moving the window, keep track of the highest sum you've seen so far.
After examining every group of numbers, divide the highest sum by the size of the group to get the maximum average.

Code Implementation

def find_max_average(numbers, subarray_length):
    current_subarray_sum = sum(numbers[:subarray_length])
    max_subarray_sum = current_subarray_sum

    # Start sliding window after initial subarray
    for i in range(subarray_length, len(numbers)):
        current_subarray_sum += numbers[i] - numbers[i - subarray_length]

        # Keep track of the largest subarray sum
        max_subarray_sum = max(max_subarray_sum, current_subarray_sum)

    # Calculating average after iterating saves time
    maximum_average = max_subarray_sum / subarray_length
    return maximum_average

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return 0 or throw an IllegalArgumentException as there's no subarray.
k is larger than the array size	Return 0 or throw an IllegalArgumentException as a subarray of size k is impossible.
Array size equals k (subarray is the whole array)	Calculate the average of the entire array.
Array contains large positive and negative numbers leading to potential overflow when summing	Use long type to store the sum to avoid integer overflow.
k is 1 (single element subarray)	Return the maximum element in the array, as it will have the highest average.
All elements in the array are negative	The algorithm should correctly identify the subarray with the least negative (closest to zero) average.
All elements in the array are zero	The average will be zero regardless of the subarray, so return 0.
Floating-point precision issues with very large or small average values	Consider using double for intermediate calculations and result to minimize precision loss.