Number of Distinct Averages

Easy

Asked by:

4 views

Topics:

ArraysTwo Pointers

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

As long as nums is not empty, you must repetitively:

Find the minimum number in nums and remove it.
Find the maximum number in nums and remove it.
Calculate the average of the two removed numbers.

The average of two numbers a and b is (a + b) / 2.

For example, the average of 2 and 3 is (2 + 3) / 2 = 2.5.

Return the number of distinct averages calculated using the above process.

Note that when there is a tie for a minimum or maximum number, any can be removed.

Example 1:

Input: nums = [4,1,4,0,3,5]
Output: 2
Explanation:
1. Remove 0 and 5, and the average is (0 + 5) / 2 = 2.5. Now, nums = [4,1,4,3].
2. Remove 1 and 4. The average is (1 + 4) / 2 = 2.5, and nums = [4,3].
3. Remove 3 and 4, and the average is (3 + 4) / 2 = 3.5.
Since there are 2 distinct numbers among 2.5, 2.5, and 3.5, we return 2.

Example 2:

Input: nums = [1,100]
Output: 1
Explanation:
There is only one average to be calculated after removing 1 and 100, so we return 1.

Constraints:

2 <= nums.length <= 100
nums.length is even.
0 <= nums[i] <= 100

Try coding it in LeetCode

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 range of values for the numbers in the input array? Can they be negative, zero, or floating-point numbers?
Can the input array be empty or null? If so, what should I return in those cases?
Are duplicate numbers allowed in the input array, and how should they be handled when calculating averages?
What constitutes a "distinct" average? Are two averages considered distinct if they are within a certain tolerance of each other (in the case of floating point numbers)?
If the number of elements in the input array is odd, should I ignore the middle element or is there an expected behavior?

Brute Force Solution

Approach

The brute force approach to finding the distinct averages involves checking every possible pair of numbers. We'll find the average of each pair and then remove any duplicate averages we find.

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

First, we'll take the smallest and largest numbers from the collection.
Then, we'll compute their average.
Next, we'll pick the next smallest and next largest numbers, calculate their average.
We continue picking pairs and calculating their averages until we've used every number.
As we find each average, we need to remember if we've seen that average before.
If the average is new, we remember it; otherwise, we ignore it because we only want unique averages.
Finally, we count how many unique averages we found during the entire process.

Code Implementation

def number_of_distinct_averages(numbers):
    numbers.sort()
    left_index = 0
    right_index = len(numbers) - 1
    distinct_averages = set()

    # Iterate while the left and right indices haven't met
    while left_index < right_index:
        average = (numbers[left_index] + numbers[right_index]) / 2

        # Add the average to the set of distinct averages
        # if it's not already present
        if average not in distinct_averages:
            distinct_averages.add(average)

        left_index += 1
        right_index -= 1

    # Return the count of distinct averages
    return len(distinct_averages)

Big(O) Analysis

Time Complexity

O(n log n) – Sorting the input array of size n takes O(n log n) time. We then iterate through the sorted array, pairing the smallest and largest elements in each step. This pairing and averaging process involves iterating through approximately n/2 pairs. Since the sorting dominates the pairing and averaging steps, the overall time complexity is O(n log n).

Space Complexity

O(N) – The space complexity is determined by the need to remember the unique averages encountered. As described in the steps, we keep track of the averages we have seen before, adding them to a data structure if they are new. In the worst-case scenario, where all calculated averages are distinct, the size of this data structure (e.g., a set or list) will grow linearly with the input size, N, where N is the number of elements in the input array. Therefore, the auxiliary space used is proportional to N.

Optimal Solution

Approach

To efficiently find the number of distinct averages, we will first organize the numbers. Then, we'll use a clever pairing strategy to quickly compute and track the unique averages, avoiding redundant calculations.

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

First, arrange all the numbers in increasing order from smallest to largest.
Take the smallest and largest numbers and find their average.
Keep track of this average in a special place (like a list or a set) so you remember it.
Remove the smallest and largest numbers you just used from the original list.
Repeat the process: find the average of the new smallest and largest numbers, remember it, and remove those numbers from your list.
Keep going until there are no numbers left in your list.
Finally, count how many different averages you recorded in your special place. This count is the answer.

Code Implementation

def number_of_distinct_averages(numbers):
    numbers.sort()
    distinct_averages = set()
    left_index = 0
    right_index = len(numbers) - 1

    # Iterate until all pairs are processed.
    while left_index < right_index:
        average = (numbers[left_index] + numbers[right_index]) / 2
        distinct_averages.add(average)

        # Move indices to the next smallest and largest elements.
        left_index += 1
        right_index -= 1

    # The number of elements in the set gives the answer
    return len(distinct_averages)

Big(O) Analysis

Time Complexity

O(n log n) – Sorting the input array of size n takes O(n log n) time. The subsequent while loop iterates at most n/2 times, where n is the number of elements in the input array. Inside the loop, finding the minimum and maximum elements (after removing the previous ones) *could* take O(n) time in each iteration, but since the array is sorted, accessing the smallest and largest element takes O(1) time. The set insertion for tracking distinct averages takes O(1) on average. Therefore, the dominant operation is the initial sorting, giving a time complexity of O(n log n).

Space Complexity

O(N) – The algorithm uses a set to store the distinct averages. In the worst-case scenario, each pair of smallest and largest numbers could result in a unique average. This would mean that the set could potentially store up to N/2 distinct averages, where N is the number of input numbers. Therefore, the auxiliary space required to store the distinct averages grows linearly with the input size, resulting in a space complexity of O(N).

Edge Cases

Case	How to Handle
Empty input array	Return 0 because there are no averages to calculate.
Array with only one element	Return 0 because an average requires at least two elements.
Array with duplicate numbers	The algorithm should handle duplicates correctly as different indices of the same number will form valid pairs.
Array with a large number of elements	Ensure the time complexity is efficient (e.g., O(n log n) with sorting) and doesn't cause performance issues.
Array with all elements being the same value	The single average will be that value and the result will be 1.
Integer overflow when calculating the sum of two numbers	Use a data type that can accommodate larger values (e.g., long) or check for potential overflow before adding.
Array contains negative numbers	The algorithm should function correctly with negative numbers without any modifications.
Floating point precision issues	Compare the calculated averages using a small tolerance to account for floating point inaccuracies instead of using equality.

Empty input array

How to Handle:

Return 0 because there are no averages to calculate.

Array with only one element

How to Handle:

Return 0 because an average requires at least two elements.

Array with duplicate numbers

How to Handle:

The algorithm should handle duplicates correctly as different indices of the same number will form valid pairs.

Array with a large number of elements

How to Handle:

Ensure the time complexity is efficient (e.g., O(n log n) with sorting) and doesn't cause performance issues.

Array with all elements being the same value

How to Handle:

The single average will be that value and the result will be 1.

Integer overflow when calculating the sum of two numbers

How to Handle:

Use a data type that can accommodate larger values (e.g., long) or check for potential overflow before adding.

Array contains negative numbers

How to Handle:

The algorithm should function correctly with negative numbers without any modifications.

Floating point precision issues

How to Handle:

Compare the calculated averages using a small tolerance to account for floating point inaccuracies instead of using equality.