**π‘ Problem Formulation:** You are given an array of integers, and you need to determine if it’s possible to reorder the array such that it forms an array of doubled pairs. For example, given the array `[4, -2, 2, -4]`

, a possible reordered array that forms doubled pairs is `[-2, -4, 2, 4]`

.

## Method 1: Counting with a Hash Map

This method involves using a hash map (dictionary in Python) to count the occurrences of each element. Afterward, we iterate through the sorted keys and try to form pairs with their double. It’s an effective method because it allows us to quickly find and match pairs.

Here’s an example:

# Imports 'collections' for the Counter class. from collections import Counter def canFormDoubledArray(arr): # Count the occurrences of each element in the array. count = Counter(arr) # Sort the keys of the counter and attempt to form pairs. for x in sorted(count, key=abs): if count[x] > count[2 * x]: return False count[2 * x] -= count[x] return True arr = [4, -2, 2, -4] print(canFormDoubledArray(arr))

The output of the above code is:

True

The code snippet starts by counting the frequency of each element in the array using the `Counter`

class. It then sorts the keys (the elements) by their absolute values to ensure we pair negative numbers with their doubles correctly. The function returns `False`

as soon as it finds an element that cannot be paired with its double, otherwise, it returns `True`

.

## Method 2: Greedy Pairing with Sorting

This method sorts the array and greedily pairs each element with its double if possible. It’s a straightforward approach that works well when the array is not too large, and elements are not excessively repeated.

Here’s an example:

def canFormDoubledArray(arr): arr.sort() while arr: value = arr.pop() if value / 2 in arr: arr.remove(value / 2) else: return False return True arr = [4, 1, 2, 8] print(canFormDoubledArray(arr))

The output of the above code is:

True

After sorting the array, the code uses a while loop to pop elements from the end of the array and tries to find their half in the remaining array. If the half is found, it’s removed from the list; otherwise, the function returns `False`

. If all elements can be paired, `True`

is returned.

## Method 3: MultiSet with Ordered Elements

Utilizing a multiset (like `collections.Counter`

in Python), which is a set that allows multiple occurrences of elements, pairs are removed starting from the smallest absolute value. This prevents pairing issues with negative numbers and zeroes.

Here’s an example:

from collections import Counter def canFormDoubledArray(arr): count = Counter(arr) for x in sorted(arr, key=abs): if count[x] == 0: continue if count[2 * x] == 0: return False count[x] -= 1 count[2 * x] -= 1 return True arr = [0, 0, 3, 6, -3, -6] print(canFormDoubledArray(arr))

The output of the above code is:

True

The code counts the occurrences of elements and then sorts the original array by their absolute values. It iterates over the sorted array, decrementing the count of each element and its double while making sure that none of them runs out before the pairing is complete. If all pairings are successful, we return `True`

.

## Method 4: Using a Set for Validation

This method involves populating a set with all elements from the array and attempting to find a doubled match for the smallest element repeatedly until the set is empty.

Here’s an example:

def canFormDoubledArray(arr): unique_elements = set(arr) arr.sort() for x in arr: if 2 * x in unique_elements: unique_elements.discard(x) unique_elements.discard(2 * x) return len(unique_elements) == 0 arr = [1, 2, 4, 8] print(canFormDoubledArray(arr))

The output of the above code is:

True

In this code snippet, a set of unique elements is created and the array is sorted. The algorithm iteratively checks if the double of the current element is present in the set. If it is, both elements are discarded from the set. The function returns `True`

if the set is empty afterward, indicating all elements were paired.

## Bonus One-Liner Method 5: Using List Comprehension and All

For concise code enthusiasts, this one-liner checks all elements to see if their double or half (if it’s even) is present in the array, after sorting it. This method sacrifices readability for brevity.

Here’s an example:

arr = sorted([1, 2, 4, 8], key=abs) print(all(arr.count(element) == arr.count(element * 2) for element in set(arr)))

The output of the above code is:

True

This one-liner sorts the array by absolute value and then uses a combination of `all()`

, `set()`

, and list comprehension to ensure that the count of each element in the array is equal to the count of its double, resulting in a quick validation of the pairing condition.

## Summary/Discussion

**Method 1: Counting with a Hash Map.**Thoroughly manages frequencies and handles negatives well. However, it may not be the most intuitive at first glance.**Method 2: Greedy Pairing with Sorting.**Simple and easy to understand. It becomes inefficient when the array has many duplicated elements.**Method 3: MultiSet with Ordered Elements.**It covers edge cases effectively and works well with zero and negative elements. It’s a bit more complex and also requires sorting.**Method 4: Using a Set for Validation.**Avoids additional space for counting but relies on sorting and set operations, which may be slower on larger arrays.**Method 5: Bonus One-Liner.**It’s the most concise method, but it’s not efficient due to repeated counting for every element and lack of readability.