**π‘ Problem Formulation:** Imagine you have two arrays: the original array and a duplicated version from which one element is missing. The challenge is to efficiently identify that lost element. For instance, if `original = [1, 2, 3, 4]`

and `duplicated = [2, 3, 1]`

, the method should return `4`

as it’s the missing element.

## Method 1: Use Summation

This approach leverages the arithmetic progression sum formula. In Python, we calculate the sum of elements in both arrays and find the difference. The difference is the lost element. This is a quick and efficient method, particularly when dealing with small to medium-size arrays.

Here’s an example:

def find_lost_element(original, duplicated): return sum(original) - sum(duplicated) original = [1, 2, 3, 4] duplicated = [2, 3, 1] print(find_lost_element(original, duplicated))

Output: `4`

In this snippet, the `find_lost_element`

function computes the sum of the `original`

array and subtracts the sum of the `duplicated`

array to identify the missing element.

## Method 2: XOR Operation

Using bitwise XOR, we exploit the property that `x ^ x = 0`

and `x ^ 0 = x`

. Hence, XOR-ing all elements in both arrays will cancel out the duplicates and leave the lost element. This method efficiently handles larger amounts of data, given its constant space complexity.

Here’s an example:

def find_lost_element_xor(original, duplicated): result = 0 for num in original + duplicated: result ^= num return result original = [1, 2, 3, 4] duplicated = [2, 3, 1] print(find_lost_element_xor(original, duplicated))

Output: `4`

In this code, the `find_lost_element_xor`

function initializes a result variable to 0 and then traverses both arrays, applying the XOR operation. The final result is the lost element.

## Method 3: Use Collections

This method leverages Python’s `collections.Counter`

class to count occurrences of elements in both arrays. By comparing the two counts, the missing element reveals itself. It’s a simple, readable approach but may have slightly slower performance on large arrays.

Here’s an example:

from collections import Counter def find_lost_element_counter(original, duplicated): original_count = Counter(original) duplicated_count = Counter(duplicated) for elem in original_count: if original_count[elem] != duplicated_count[elem]: return elem original = [1, 2, 3, 4] duplicated = [2, 3, 1] print(find_lost_element_counter(original, duplicated))

Output: `4`

Here, `find_lost_element_counter`

makes two counts and compares them to detect the element which occurs less frequently in the `duplicated`

array, identifying the lost element.

## Method 4: Use Set Difference

The set difference operation is useful here. When converting arrays to sets, we can easily find the lost element as sets contain unique items only. This method is extremely simple and works well with unsorted arrays.

Here’s an example:

def find_lost_element_set(original, duplicated): return (set(original) - set(duplicated)).pop() original = [1, 2, 3, 4] duplicated = [2, 3, 1] print(find_lost_element_set(original, duplicated))

Output: `4`

This `find_lost_element_set`

function converts both arrays to sets and then subtracts them, with the `pop()`

method retrieving the lost element from the resulting set.

## Bonus One-Liner Method 5: List Comprehension

A Pythonic one-liner using list comprehension can also solve this problem. It compares elements’ occurrences directly, thus identifying the lost one. It is concise and elegant but may be less efficient on larger arrays.

Here’s an example:

original = [1, 2, 3, 4] duplicated = [2, 3, 1] lost_element = [x for x in original if original.count(x) != duplicated.count(x)][0] print(lost_element)

Output: `4`

This one-liner traverses the original array once and checks the count of each element against the duplicated one, appending the lost one to the list and selecting the first entry.

## Summary/Discussion

**Method 1:**Summation. Simple and efficient for small arrays. Not ideal for very large arrays due to potential overflow.**Method 2:**XOR Operation. Extremely efficient with constant space complexity. Requires understanding of bitwise operators.**Method 3:**Collections. Intuitive and easy to read code. May not be as performant for very large datasets due to counting overhead.**Method 4:**Set Difference. Excellent for unsorted arrays and readability. Will not work properly if the arrays have multiple duplicates of elements.**Bonus Method 5:**List Comprehension. Pythonic and concise. Performance could degrade with larger arrays due to multiple passes for counting.