Python – Find Kth Even Element

Finding the Kth even element in a list is a common programming task. Python provides several approaches using list comprehension, filtering, and indexing techniques.

Method 1: Using List Comprehension

Create a list of even elements and access the Kth element directly ?

numbers = [14, 63, 28, 32, 18, 99, 13, 61]

print("The list is:")
print(numbers)

K = 3
print("The value of K is:")
print(K)

# Find Kth even element (1-indexed)
result = [element for element in numbers if element % 2 == 0][K-1]

print("The Kth even element is:")
print(result)

The list is:
[14, 63, 28, 32, 18, 99, 13, 61]
The value of K is:
3
The Kth even element is:
28

Method 2: Using Filter and List Conversion

Use the built-in filter() function for cleaner code ?

numbers = [14, 63, 28, 32, 18, 99, 13, 61]
K = 2

even_numbers = list(filter(lambda x: x % 2 == 0, numbers))
print("Even numbers:", even_numbers)

kth_even = even_numbers[K-1]
print(f"The {K}th even element is: {kth_even}")

Even numbers: [14, 28, 32, 18]
The 2nd even element is: 28

Method 3: Iterative Approach with Counter

Use a counter to find the Kth even element without creating intermediate lists ?

def find_kth_even(numbers, k):
    count = 0
    for num in numbers:
        if num % 2 == 0:
            count += 1
            if count == k:
                return num
    return None

numbers = [14, 63, 28, 32, 18, 99, 13, 61]
K = 4

result = find_kth_even(numbers, K)
if result is not None:
    print(f"The {K}th even element is: {result}")
else:
    print(f"There are fewer than {K} even elements in the list")

The 4th even element is: 18

Comparison

Conclusion

Use list comprehension for simple cases with small datasets. For memory efficiency with large lists, prefer the iterative approach that stops once the Kth even element is found.