5 Best Ways to Find the Number Occurring an Odd Number of Times Using Lambda and Reduce in Python

💡 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, it’s 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.