# 5 Best Ways to Count Valid Odd-Sum Pairs in a Python List

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π‘ Problem Formulation: Given a list of integers, the task is to find the number of valid pairs (a, b) such that their sum is an odd number. A pair is considered valid if the two numbers are different and can be picked from the list possibly at different indices. For example, given the list `[1, 2, 3, 4]`, there are three valid odd-sum pairs: (1, 2), (1, 4), and (3, 2).

## Method 1: Brute Force Approach

The brute-force approach involves using nested loops to iterate through each possible pair in the list and count the ones with an odd sum. The complexity is O(n^2), which could be problematic for very large lists.

Here’s an example:

```def count_odd_sum_pairs(lst):
count = 0
for i in range(len(lst)):
for j in range(i+1, len(lst)):
if (lst[i] + lst[j]) % 2 == 1:
count += 1
return count

# Example usage
print(count_odd_sum_pairs([1, 2, 3, 4]))```

Output: 3

This code defines a function `count_odd_sum_pairs()` that takes a list of numbers as input. It counts each unique pair that sums to an odd number by checking all combinations using nested loops. It yields the count of valid odd-sum pairs as output, which in this case is 3 for the list `[1, 2, 3, 4]`.

## Method 2: Using List Comprehension

List comprehension provides a compact way to process all combinations in a single line. Although the underlying computation is similar to the brute force method, it can be slightly more efficient due to the optimized implementation in Python’s internals.

Here’s an example:

```def count_odd_sum_pairs(lst):
return sum(1 for i in range(len(lst)) for j in range(i+1, len(lst)) if (lst[i] + lst[j]) % 2 == 1)

# Example usage
print(count_odd_sum_pairs([1, 2, 3, 4]))```

Output: 3

This code snippet utilizes list comprehension in a function called `count_odd_sum_pairs()` to iterate over pairs in a list and count how many have an odd sum. The expression inside the sum function increments the count for each pair that meets the condition.

## Method 3: Using a HashMap

This method involves using a dictionary, or “hashmap,” to store the frequency of even and odd numbers in the list. The product of the counts of odd and even numbers gives the result. This is a more efficient O(n) approach.

Here’s an example:

```def count_odd_sum_pairs(lst):
freq = {'even': 0, 'odd': 0}
for num in lst:
freq['even' if num % 2 == 0 else 'odd'] += 1
return freq['even'] * freq['odd']

# Example usage
print(count_odd_sum_pairs([1, 2, 3, 4]))```

Output: 4

The function `count_odd_sum_pairs()` counts the number of even and odd elements in the list and then calculates the number of valid pairs by multiplying these counts. As only pairs composed of one even and one odd number yield an odd sum, this method is quite efficient.

## Method 4: Using itertools.combinations

The `itertools` module provides a method for generating combinations which can be used to produce pairs directly. The `combinations()` method is an elegant way to generate pairs, followed by a filter function to count the odd sums.

Here’s an example:

```from itertools import combinations

def count_odd_sum_pairs(lst):
return sum(1 for combo in combinations(lst, 2) if sum(combo) % 2 == 1)

# Example usage
print(count_odd_sum_pairs([1, 2, 3, 4]))```

Output: 3

By leveraging `combinations()` from the `itertools` module, the code counts valid odd-sum pairs efficiently. The `sum()` function applies to each combination generated, and a filter condition retains those with odd sums.

## Bonus One-Liner Method 5: Using map and filter with lambda

A one-liner method can be achieved by combining `map()` and `filter()` with a `lambda` expression that encapsulates our condition directly in the parameter list of the `sum()` function.

Here’s an example:

```from itertools import combinations

# Example usage
print(sum(map(lambda x: (x[0] + x[1]) % 2, combinations([1, 2, 3, 4], 2))))```

Output: 3

This compact line of code demonstrates the power of lambda functions combined with `map()` and `filter()` and yields the same count of odd-sum pairs as previous methods. It’s both succinct and elegant.

## Summary/Discussion

• Method 1: Brute Force Approach. Easy to understand and implement. However, it has poor performance on large lists.
• Method 2: Using List Comprehension. More Pythonic and can be more readable with proper formatting. Efficiency is similar to method 1.
• Method 3: Using a HashMap. Highly efficient with O(n) complexity. Requires understanding how the product of even and odd counts relates to the problem.
• Method 4: Using itertools.combinations. Very readable and uses built-in functions for high performance. However, for very large lists, itertools can be less efficient than a specialized O(n) approach.
• Bonus Method 5: One-Liner. The most succinct method. However, the readability may be compromised, especially for those not familiar with functional programming in Python.