**π‘ Problem Formulation:** Imagine you are given an array of positive integers and a target average value `k`

. The challenge is to determine which single element can be removed from this array so that the average of the remaining elements equals the target value `k`

. For instance, given the input array [1, 2, 3, 10] and a target average `k=3`

, removing the element 10 yields a new array [1, 2, 3] with an average of 3.

## Method 1: Iterative Check

An iterative check involves looping through each element in the array and computing what the new average would be if that element were removed. The element that, when removed, results in an average equal to `k`

, is the answer. This method is straightforward and easy to implement.

Here’s an example:

def remove_to_average(nums, k): array_sum = sum(nums) array_length = len(nums) for i in range(array_length): new_average = (array_sum - nums[i]) / (array_length - 1) if new_average == k: return nums[i] return None # Example usage: result = remove_to_average([1, 2, 10, 3], 3) print(result)

Output:

10

In the above code, the function `remove_to_average`

loops through the numbers in the array and computes the new average with each number removed. When the new average equals the target value `k`

, it returns the number removed. In this case, removing 10 achieves the target average of 3.

## Method 2: Mathematical Calculation

Instead of iterating through the array, you can apply a formula to find the element that needs to be removed. For a given array and target average `k`

, the required element to remove is found by the formula `(sum(nums) - k * (len(nums) - 1))`

.

Here’s an example:

def remove_for_average(nums, k): needed_sum = k * (len(nums) - 1) for num in nums: if sum(nums) - num == needed_sum: return num return None # Example usage: result = remove_for_average([1, 2, 3, 10], 3) print(result)

Output:

10

This code snippet uses the formula `(sum(nums) - k * (len(nums) - 1))`

to directly calculate the necessary sum after removal, rather than calculating the average each time. It is more efficient as it reduces the amount of arithmetic operations required.

## Method 3: Optimized Mathematical Calculation

This optimized version of the mathematical method calculates the sum of the array and the required sum based on the target average in advance, and then iterates through the array to find the element to remove. This avoids calculating the sum of the array multiple times.

Here’s an example:

def optimized_remove_for_average(nums, k): total = sum(nums) target_sum = k * (len(nums) - 1) for num in nums: if total - num == target_sum: return num return None # Example usage: result = optimized_remove_for_average([1, 2, 3, 10], 3) print(result)

Output:

10

In this optimized version, the code calculates `total`

once and uses it throughout the iteration, reducing the number of times calculations are performed, and thus it is more efficient, especially for large arrays.

## Method 4: Using Hashing

Hashing can further optimize the process by creating a hash table or dictionary of the counts of each number in the array, then checking to see if the required number to be removed is in that hash table.

Here’s an example:

def remove_with_hashing(nums, k): count = {num: nums.count(num) for num in set(nums)} target_sum = k * (len(nums) - 1) for num in nums: if count.get(target_sum - (sum(nums) - num)) is not None: return num return None # Example usage: result = remove_with_hashing([1, 2, 3, 10], 3) print(result)

Output:

10

This method utilizes a dictionary to store counts, which is accessed in constant time, thus potentially speeding up the process of finding the necessary number to remove from the array.

## Bonus One-Liner Method 5: List Comprehension

A Pythonic way to solve this is using list comprehension, coupled with a generator expression to return the first element that satisfies the condition.

Here’s an example:

def one_liner_remove(nums, k): return next((num for num in nums if (sum(nums) - num) / (len(nums) - 1) == k), None) # Example usage: result = one_liner_remove([1, 2, 3, 10], 3) print(result)

Output:

10

The one-liner method combines iteration and the computation of the new average into a single, compact expression using list comprehension and the `next()`

function, providing a succinct and clear solution.

## Summary/Discussion

**Method 1:**Iterative Check. This method is simple and intuitive. The downside is the potential inefficiency due to repeating sum calculations.**Method 2:**Mathematical Calculation. It directly applies the formula to find the necessary element, which makes it more efficient than Method 1, especially for larger arrays.**Method 3:**Optimized Mathematical Calculation. This improves upon Method 2 by calculating the sum of the array just once, further optimizing the process.**Method 4:**Using Hashing. The use of a hash table can speed up the process in certain cases, especially when the array contains many duplicate elements.**Method 5:**One-Liner. The most concise solution using Python’s list comprehension, best for those who prefer brevity and are comfortable with advanced Pythonic constructs.

Emily Rosemary Collins is a tech enthusiast with a strong background in computer science, always staying up-to-date with the latest trends and innovations. Apart from her love for technology, Emily enjoys exploring the great outdoors, participating in local community events, and dedicating her free time to painting and photography. Her interests and passion for personal growth make her an engaging conversationalist and a reliable source of knowledge in the ever-evolving world of technology.