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Minimum Average Difference

Medium

Amazon

1 view

Topics:

Arrays

You are given a 0-indexed integer array nums of length n.

The average difference of the index i is the absolute difference between the average of the first i + 1 elements of nums and the average of the last n - i - 1 elements. Both averages should be rounded down to the nearest integer.

Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.

Note:

The absolute difference of two numbers is the absolute value of their difference.
The average of n elements is the sum of the n elements divided (integer division) by n.
The average of 0 elements is considered to be 0.

Example 1:

Input: nums = [2,5,3,9,5,3]
Output: 3

Example 2:

Input: nums = [0]
Output: 0

Can you provide an algorithm to solve this problem efficiently?

Minimum Average Difference

Medium

Amazon

1 view

Topics:

Arrays

You are given a 0-indexed integer array nums of length n.

Return the index with the minimum average difference. If there are multiple such indices, return the smallest one.

Note:

The absolute difference of two numbers is the absolute value of their difference.
The average of n elements is the sum of the n elements divided (integer division) by n.
The average of 0 elements is considered to be 0.

Example 1:

Input: nums = [2,5,3,9,5,3]
Output: 3

Example 2:

Input: nums = [0]
Output: 0

Can you provide an algorithm to solve this problem efficiently?

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 are the constraints on the size of the input array `nums`? What is the maximum possible value of an element in `nums`?
Can the input array `nums` contain negative numbers, zeros, or floating-point numbers, or will it always contain positive integers?
What should I return if the input array is empty or null?
If there are multiple indices that result in the same minimum average difference, should I return the smallest such index?
Is the average difference calculated using integer division (truncating the decimal portion) or should I round the result to the nearest integer?

Brute Force Solution

Approach

The brute force strategy for finding the minimum average difference involves checking every single possible division point in the list. We'll calculate the average difference for each possible division, and then find the smallest one.

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

First, we consider dividing the list after the very first number.
We calculate the average of the numbers before this division point, and the average of the numbers after this division point.
We find the difference between these two averages.
Then, we consider dividing the list after the second number, and repeat the process of calculating the two averages and their difference.
We keep doing this, moving the division point one number at a time, until we've tried every possible division point in the list.
Finally, after checking all the possible division points and calculating all the average differences, we select the smallest difference we found.

Code Implementation

def minimum_average_difference_brute_force(numbers):
    list_length = len(numbers)
    minimum_average_difference = float('inf')
    minimum_average_difference_index = -1

    for current_index in range(list_length):
        # Calculate the average of the left side of the array
        left_side_sum = sum(numbers[0:current_index+1])
        left_side_length = current_index + 1
        left_side_average = left_side_sum // left_side_length

        # Calculate the average of the right side of the array
        if current_index == list_length - 1:
            right_side_average = 0
        else:
            right_side_sum = sum(numbers[current_index+1:list_length])
            right_side_length = list_length - current_index - 1
            right_side_average = right_side_sum // right_side_length

        # Comparing the absolute difference to minimum value
        average_difference = abs(left_side_average - right_side_average)

        if average_difference < minimum_average_difference:
            # Track the lowest absolute average difference
            minimum_average_difference = average_difference
            minimum_average_difference_index = current_index

    return minimum_average_difference_index

Big(O) Analysis

Time Complexity

O(n) – The provided brute force strategy iterates through the array of n elements to determine each possible division point. For each division point, it calculates the average of the numbers before and after the point. Calculating these averages involves summing a subset of the array elements. In the original prompt, it implied the subset summation was done repeatedly in a nested way which gives O(n^2), but actually it can be precalculated and be done in O(n) with prefix sums. For each index, we can get the prefix sum to get the average of the first part and then calculate the second part from the sum of the array. Therefore the overall operations are proportional to n.

Space Complexity

O(1) – The provided plain English explanation outlines an iterative approach that calculates averages and differences without storing intermediate division results. It only requires a few constant-sized variables to keep track of the current division point, the current averages, and the minimum average difference found so far. The space used doesn't scale with the input size N (the number of elements in the list). Therefore, the auxiliary space complexity is O(1).

Optimal Solution

Approach

To find the minimum average difference, we want to efficiently calculate the average difference for every possible split point in the data. The key is to avoid recomputing sums repeatedly by keeping track of the running total from both the beginning and the end.

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

First, calculate the total sum of all the numbers.
Then, as you move through the numbers one by one, keep track of the running sum from the beginning.
At each step, calculate the average of the numbers from the beginning up to the current number.
Also, calculate the average of the remaining numbers from the current number to the end by subtracting the running sum from the total sum and dividing by the count of the remaining numbers. Be careful with division by zero if there are no remaining numbers.
Find the absolute difference between these two averages.
Keep track of the minimum of all these absolute differences that you've found so far.
After going through all the numbers, the minimum absolute difference you've tracked will be the answer.

Code Implementation

def minimum_average_difference(numbers):
    total_sum = sum(numbers)
    current_sum = 0
    minimum_average_difference_so_far = float('inf')
    minimum_index = 0

    for index, number in enumerate(numbers):
        current_sum += number

        # Calculate average of the first part of the array.
        first_part_average = current_sum // (index + 1)

        # Calculate average of the second part, handling division by zero.
        if index == len(numbers) - 1:
            second_part_average = 0
        else:
            second_part_average = (total_sum - current_sum) // (len(numbers) - index - 1)

        # Get the absolute difference between the two averages
        absolute_difference = abs(first_part_average - second_part_average)

        # Keep track of the min average difference
        if absolute_difference < minimum_average_difference_so_far:

            minimum_average_difference_so_far = absolute_difference
            minimum_index = index

    return minimum_index

Big(O) Analysis

Time Complexity

O(n) – The algorithm iterates through the input array nums of size n only once to calculate prefix sums and averages. Within the loop, calculations such as calculating prefix sums, calculating the average of the first part of the array and the average of the remaining part, and computing absolute differences, are all constant-time operations O(1). Therefore, since the loop runs n times and each loop takes O(1) time, the total time complexity is O(n).

Space Complexity

O(1) – The algorithm calculates running sums and averages in place, primarily using a few variables to store the total sum, current running sum, minimum average difference, and index. These variables consume a constant amount of memory regardless of the input array's size (N). Therefore, the auxiliary space complexity is O(1).

Edge Cases

Case	How to Handle
Empty input array	Return -1 or throw an exception as the average difference is undefined for an empty array.
Single element array	The average difference is the absolute value of the single element, divided by one and zero, returning index 0.
Array with all elements being zero	The minimum average difference will be zero, and the index will be zero.
Array with all elements being the maximum integer value	Handle potential integer overflow when calculating prefix sums by using long data type.
Array with a mix of very large positive and very large negative numbers	Handle potential integer overflow or underflow when calculating sums using long data type and carefully consider integer representation boundaries.
Maximum sized input array (e.g., 10^5 elements)	Ensure the solution has a time complexity of O(n) to avoid exceeding time limits.
Array where the first few elements have a very large average difference, and the last few have a small one	The algorithm should correctly iterate through all possible splits of the array, maintaining an accurate minimum average difference.
Array containing duplicate numbers that result in the same average difference	The problem statement specifies returning the smallest index, so the algorithm should only update the result when a smaller average difference is encountered.

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 are the constraints on the size of the input array `nums`? What is the maximum possible value of an element in `nums`?
Can the input array `nums` contain negative numbers, zeros, or floating-point numbers, or will it always contain positive integers?
What should I return if the input array is empty or null?
If there are multiple indices that result in the same minimum average difference, should I return the smallest such index?
Is the average difference calculated using integer division (truncating the decimal portion) or should I round the result to the nearest integer?

Brute Force Solution

Approach

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

First, we consider dividing the list after the very first number.
We calculate the average of the numbers before this division point, and the average of the numbers after this division point.
We find the difference between these two averages.
Then, we consider dividing the list after the second number, and repeat the process of calculating the two averages and their difference.
We keep doing this, moving the division point one number at a time, until we've tried every possible division point in the list.
Finally, after checking all the possible division points and calculating all the average differences, we select the smallest difference we found.

Code Implementation

def minimum_average_difference_brute_force(numbers):
    list_length = len(numbers)
    minimum_average_difference = float('inf')
    minimum_average_difference_index = -1

    for current_index in range(list_length):
        # Calculate the average of the left side of the array
        left_side_sum = sum(numbers[0:current_index+1])
        left_side_length = current_index + 1
        left_side_average = left_side_sum // left_side_length

        # Calculate the average of the right side of the array
        if current_index == list_length - 1:
            right_side_average = 0
        else:
            right_side_sum = sum(numbers[current_index+1:list_length])
            right_side_length = list_length - current_index - 1
            right_side_average = right_side_sum // right_side_length

        # Comparing the absolute difference to minimum value
        average_difference = abs(left_side_average - right_side_average)

        if average_difference < minimum_average_difference:
            # Track the lowest absolute average difference
            minimum_average_difference = average_difference
            minimum_average_difference_index = current_index

    return minimum_average_difference_index

Big(O) Analysis

Time Complexity

Space Complexity

Optimal Solution

Approach

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

First, calculate the total sum of all the numbers.
Then, as you move through the numbers one by one, keep track of the running sum from the beginning.
At each step, calculate the average of the numbers from the beginning up to the current number.
Also, calculate the average of the remaining numbers from the current number to the end by subtracting the running sum from the total sum and dividing by the count of the remaining numbers. Be careful with division by zero if there are no remaining numbers.
Find the absolute difference between these two averages.
Keep track of the minimum of all these absolute differences that you've found so far.
After going through all the numbers, the minimum absolute difference you've tracked will be the answer.

Code Implementation

def minimum_average_difference(numbers):
    total_sum = sum(numbers)
    current_sum = 0
    minimum_average_difference_so_far = float('inf')
    minimum_index = 0

    for index, number in enumerate(numbers):
        current_sum += number

        # Calculate average of the first part of the array.
        first_part_average = current_sum // (index + 1)

        # Calculate average of the second part, handling division by zero.
        if index == len(numbers) - 1:
            second_part_average = 0
        else:
            second_part_average = (total_sum - current_sum) // (len(numbers) - index - 1)

        # Get the absolute difference between the two averages
        absolute_difference = abs(first_part_average - second_part_average)

        # Keep track of the min average difference
        if absolute_difference < minimum_average_difference_so_far:

            minimum_average_difference_so_far = absolute_difference
            minimum_index = index

    return minimum_index

Big(O) Analysis

Time Complexity

Space Complexity

Edge Cases

Case	How to Handle
Empty input array	Return -1 or throw an exception as the average difference is undefined for an empty array.
Single element array	The average difference is the absolute value of the single element, divided by one and zero, returning index 0.
Array with all elements being zero	The minimum average difference will be zero, and the index will be zero.
Array with all elements being the maximum integer value	Handle potential integer overflow when calculating prefix sums by using long data type.
Array with a mix of very large positive and very large negative numbers	Handle potential integer overflow or underflow when calculating sums using long data type and carefully consider integer representation boundaries.
Maximum sized input array (e.g., 10^5 elements)	Ensure the solution has a time complexity of O(n) to avoid exceeding time limits.
Array where the first few elements have a very large average difference, and the last few have a small one	The algorithm should correctly iterate through all possible splits of the array, maintaining an accurate minimum average difference.
Array containing duplicate numbers that result in the same average difference	The problem statement specifies returning the smallest index, so the algorithm should only update the result when a smaller average difference is encountered.