**π‘ Problem Formulation:** Your goal is to identify a number that appears an odd number of times within a list. For instance, given the list `[4, 3, 4, 4, 7, 3]`

, the desired output is `7`

since it’s the only number occurring an odd number of times.

## Method 1: Use Reduce with XOR Operation

This method applies the reduce function along with a lambda function performing the XOR operation to find the odd-occurring number. The XOR of two identical numbers is 0, and the XOR of a number with 0 gives the number itself, which works perfectly for finding the odd occurrence in a pair-wise cancellation.

Here’s an example:

from functools import reduce nums = [4, 3, 4, 4, 7, 3] result = reduce(lambda x, y: x ^ y, nums) print(result)

Output: `7`

This code snippet initializes a list called `nums`

, then calls `reduce()`

to apply a lambda function performing the XOR operation between elements. The XOR operation `^`

ensures that all numbers occurring an even number of times cancel out, leaving the odd-occurring number as the result.

## Method 2: Using Counter from Collections

The Counter class from the ‘collections’ module provides a simple way to count occurrences. After initializing a Counter, a loop can find the first number occurring an odd number of times.

Here’s an example:

from collections import Counter nums = [4, 3, 4, 4, 7, 3] counter = Counter(nums) for num, count in counter.items(): if count % 2 != 0: result = num break print(result)

Output: `7`

The code creates a `Counter`

object with the counts of each number. It then iterates through the counted items and checks if the count is odd. Upon finding the first odd-occurring number, it breaks out of the loop and prints the result.

## Method 3: Use Filter and Lambda to Remove Even-Occurring Numbers

This method involves filtering out all numbers that occur an even number of times, leaving only the number(s) that occur an odd number of times. It uses `filter`

and a `lambda`

function in combination with `nums.count`

method.

Here’s an example:

nums = [4, 3, 4, 4, 7, 3] result = list(filter(lambda x: nums.count(x) % 2 != 0, set(nums))) print(result[0])

Output: `7`

The example code converts the list to a set to remove duplicates and then filters it by the count occurrence using a lambda function. The filtered list contains only numbers with odd occurrences, and we extract the first element since the problem guarantees only one such number exists.

## Method 4: Nested Loop to Count Occurrences

Using nested loops can also solve the problem. This brute-force approach manually counts occurrences of each number and finds the odd one out.

Here’s an example:

nums = [4, 3, 4, 4, 7, 3] for num in nums: if nums.count(num) % 2 != 0: result = num break print(result)

Output: `7`

The code iterates through every number in the list and uses the `count()`

method to count its occurrences. If a number’s count is odd, it assigns that number to `result`

and ceases the loop.

## Bonus One-Liner Method 5: Chain Together Lambda, Reduce, and XOR

For a concise solution, you can use a combination of `lambda`

, `reduce`

, and XOR in one line of code. This is a condensed version of Method 1.

Here’s an example:

from functools import reduce print(reduce(lambda x, y: x ^ y, [4, 3, 4, 4, 7, 3]))

Output: `7`

This one-liner example directly prints the result of performing the XOR operation across all list elements using reduce and a lambda function to find the number with an odd occurrence.

## Summary/Discussion

**Method 1:**XOR and Reduce. Fast and efficient, works well with large lists. However, it only works when there is exactly one number occurring an odd number of times.**Method 2:**Counter Object. More readable, and no need for a custom function definition. Can be slower for large datasets because it constructs an intermediate collection for counting.**Method 3:**Filter and Lambda. Less efficient since it requires multiple traversals of the list for counting, but it is concise and leverages built-in Python features neatly.**Method 4:**Nested Loops. The least efficient, brute force method that can be relatively slow for large lists and lacks elegance. However, itβs straightforward and easy to understand for beginners.**Method 5:**Lambda One-Liner. Quick and slick, this method combines the reduce function with a lambda for a highly concise solution. The same strengths and weaknesses of Method 1 apply here.