**π‘ Problem Formulation:** When working with arrays in Python, it’s often necessary to determine if a specific numerical relationship exists between elements. A common scenario is checking whether a number and its double are both present within the array. For instance, given the array [1, 3, 4, 2], the output should be True as the number 2 and its double, 4, exist in the array.

## Method 1: Brute Force Approach

This method involves a straightforward approach by checking each element in the array and searching for its double using nested loops. While this method is easy to understand, it isn’t efficient with large arrays due to its O(n^2) time complexity.

Here’s an example:

def has_double_brute_force(arr): for i in arr: if 2 * i in arr: return True return False arr = [10, 3, 5, 1, 6] print(has_double_brute_force(arr))

Output: `False`

This snippet defines a function `has_double_brute_force()`

that iterates over each element in the array and checks if its double exists in the array. If a match is found, the function returns True; otherwise, it returns False after completing the iteration.

## Method 2: Using a Set for Lookup

Improving on the brute force approach, this method uses a set to store the elements of the array. The set provides O(1) average time complexity for lookups, which makes finding a number’s double significantly faster. The overall time complexity becomes O(n).

Here’s an example:

def has_double_set(arr): seen = set() for number in arr: if number * 2 in seen or (number % 2 == 0 and number // 2 in seen): return True seen.add(number) return False arr = [5, 1, 3, 10, 20] print(has_double_set(arr))

Output: `True`

This code defines a function `has_double_set()`

that uses a set for constant time lookups. It checks each number if its double or half (if the number is even) is in the set already. If a match is found, it returns True; if no matches are found after iterating through all numbers, it returns False.

## Method 3: Sorting and Binary Search

This method involves sorting the array first and then using a binary search algorithm to identify if the double of each element exists. The time complexity for sorting is O(n log n) and for binary search is O(log n), making the overall time complexity O(n log n).

Here’s an example:

import bisect def has_double_binary_search(arr): arr.sort() for i, number in enumerate(arr): index = bisect.bisect_left(arr, number * 2, i + 1) if index < len(arr) and arr[index] == number * 2: return True return False arr = [7, 3, 1, 14] print(has_double_binary_search(arr))

Output: `True`

The `has_double_binary_search()`

function first sorts the array and then for each element, utilizes the `bisect_left()`

function from Python’s bisect module to perform efficient binary searches. If a double is located at the proper index, True is returned; otherwise, the function returns False after all elements are processed.

## Method 4: Hash Map of Frequencies

Using a hash map to record the frequency of each element can help determine if a number and its double are present. After populating the hash map, another iteration checks if a double exists for elements which are not zero.

Here’s an example:

def has_double_hash_map(arr): from collections import Counter frequency = Counter(arr) for number in arr: if number != 0 and frequency[number * 2]: return True if number == 0 and frequency[number] > 1: return True return False arr = [0, 0, 3, 6] print(has_double_hash_map(arr))

Output: `True`

The function `has_double_hash_map()`

makes use of the `Counter`

class from the collections module to count occurrences of each element. It subsequently checks for the existence of a double for each non-zero element, and separately handles the case of zero, for which we need at least two zeros in the array to return True, as zero doubled is still zero.

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

This concise one-liner uses list comprehension and the inbuilt set data structure to search the array for a number and its double simultaneously.

Here’s an example:

arr = [1, 2, 4, 8] has_double = any(n in set(arr) for n in [x * 2 for x in arr]) print(has_double)

Output: `True`

The code snippet checks if any element’s double exists in the array using a single line of Python code. By generating a list of all doubles and converting the original array to a set for O(1) lookup times, we can ascertain the presence of the double of any number.

## Summary/Discussion

**Method 1: Brute Force Approach.**Strengths: Straightforward, easy to implement. Weaknesses: Poor performance on large datasets due to O(n^2) time complexity.**Method 2: Using a Set for Lookup.**Strengths: Significantly faster than brute force, with O(n) time complexity. Weaknesses: Requires additional memory for storing elements in a set.**Method 3: Sorting and Binary Search.**Strengths: Good balance between time complexity and intuitive implementation. Weaknesses: O(n log n) time complexity due to sorting, which may not be necessary with more efficient approaches.**Method 4: Hash Map of Frequencies.**Strengths: Efficiently checks for doubles, including the edge case of zero. Weaknesses: Additional space for the hash map and slightly more complex logic compared to other methods.**Method 5: Using List Comprehension and Sets.**Strengths: Elegant and concise. Weaknesses: Creates an entire list of doubles in memory, which could be memory-intensive for large arrays.