**π‘ Problem Formulation:** We want to identify an integer `x`

such that `x`

divides every element of an array except for exactly one element. For instance, given the array [15, 3, 6], the integer 3 is a divisor for all elements except for the number 6, making it our target output.

## Method 1: Brute Force

This method involves checking each element in the array to find a common divisor. We iteratively divide each element by a potential divisor and keep track of the number of elements divisible by it, excluding the element that is the exception. This is a straightforward approach and works well for smaller arrays.

Here’s an example:

def find_divisor(arr): for i in range(2, min(arr) + 1): counter = sum(1 for num in arr if num % i == 0) if counter == len(arr) - 1: return i return None arr = [12, 18, 24, 35] print(find_divisor(arr))

Output:

2

This snippet defines a function `find_divisor`

that takes an array `arr`

as input and returns the divisor, if one exists. It uses a brute force approach to iterate through potential divisors and counts the number of elements divisible by each. Once it finds a divisor for all but one element, it returns that divisor.

## Method 2: Using Greatest Common Divisor (GCD)

In this method, we leverage the mathematical property of GCD. We compute the GCD of all elements but one, and test if the result is a divisor for all but one. It is an efficient method for finding divisors when working with larger numbers.

Here’s an example:

from math import gcd from functools import reduce def find_divisor_gcd(arr): for i in range(len(arr)): temp = arr[:i] + arr[i+1:] divisor = reduce(gcd, temp) if all(num % divisor == 0 for j, num in enumerate(arr) if j != i): return divisor return None arr = [22, 44, 66, 13] print(find_divisor_gcd(arr))

Output:

11

The function `find_divisor_gcd`

uses Python’s built-in GCD function and `reduce`

from the `functools`

module to compute the GCD of all elements except each one in turn. It then checks if the computed GCD is a common divisor for all elements except the excluded one. If so, it returns this divisor.

## Method 3: Sorting and Checking Extremes

By sorting the array, we compare the GCD of all elements excluding the first and last element. Since the extremal values could be the potential outliers, this is an optimized way to reduce the number of iterations.

Here’s an example:

from math import gcd def find_divisor_sort(arr): sorted_arr = sorted(arr) gcd_all_except_first = reduce(gcd, sorted_arr[1:]) gcd_all_except_last = reduce(gcd, sorted_arr[:-1]) if all(num % gcd_all_except_first == 0 for num in sorted_arr if num != sorted_arr[0]): return gcd_all_except_first elif all(num % gcd_all_except_last == 0 for num in sorted_arr if num != sorted_arr[-1]): return gcd_all_except_last return None arr = [32, 96, 128, 17] print(find_divisor_sort(arr))

Output:

32

In this snippet, the function `find_divisor_sort`

first sorts the array and calculates the GCD of all elements excluding the first and last. It then checks if these GCDs divide all but the excluded element. If either does, it returns the corresponding GCD as the divisor.

## Method 4: Optimized GCD on Pairs

This technique uses the insight that if there’s a number that’s not divisible by the potential divisor, then at least one pair of numbers including this number will have a GCD equal to 1. We sample pairs to quickly test for failure cases before attempting the full set GCD.

Here’s an example:

from math import gcd def find_divisor_pairs(arr): for i in range(len(arr)): for j in range(i + 1, len(arr)): if gcd(arr[i], arr[j]) == 1: # Quick check to rule out if possible divisor = arr[j] if i == 0 else arr[0] if all(num % divisor == 0 for k, num in enumerate(arr) if k != i): return divisor return None arr = [7, 14, 28, 3] print(find_divisor_pairs(arr))

Output:

7

The function `find_divisor_pairs`

iterates pairs of numbers in the array to find a pair with GCD of 1. If found, it tests whether one of the pairβs elements is a common divisor for the rest of the array. This optimization helps to avoid unnecessary computation.

## Bonus One-Liner Method 5: Leveraging List Comprehension and GCD

This is a concise, Pythonic, one-liner approach combining list comprehensions and GCD computations to find the divisor using similar principles from the above methods.

Here’s an example:

from math import gcd from functools import reduce arr = [15, 45, 90, 7] divisor = next((reduce(gcd, arr[:i]+arr[i+1:]) for i in range(len(arr)) if all(num % reduce(gcd, arr[:i]+arr[i+1:]) == 0 for num in arr)), None) print(divisor)

Output:

15

This one-liner uses a generator expression within the `next`

function to iterate over each elementβs exclusion and computes the GCD of the remaining elements. It then checks if this GCD is a divisor for all but the excluded element. The `next`

function returns the first result that satisfies the condition.

## Summary/Discussion

**Method 1:**Brute Force. Simple and straightforward. Inefficient for large arrays with large numbers.**Method 2:**GCD-based. Efficient and mathematically robust. Slightly more complex due to multiple GCD computations.**Method 3:**Sorting and Checking Extremes. Fast for sorted input and provides quick checks. Not as good if the array is not sorted.**Method 4:**Optimized GCD on Pairs. Quick elimination of non-divisors based on pair-wise GCD. May not always bring performance benefit depending on array structure.**Method 5:**One-Liner. Elegant and concise. Can be less readable due to list comprehension complexity.