Welcome to the Treehouse Community

Want to collaborate on code errors? Have bugs you need feedback on? Looking for an extra set of eyes on your latest project? Get support with fellow developers, designers, and programmers of all backgrounds and skill levels here with the Treehouse Community! While you're at it, check out some resources Treehouse students have shared here.

Looking to learn something new?

Treehouse offers a seven day free trial for new students. Get access to thousands of hours of content and join thousands of Treehouse students and alumni in the community today.

Start your free trial

Computer Science Introduction to Algorithms Algorithms in Code Linear Search in Code

Posted by Brittney Coble
Brittney Coble
16,474 Points

Getting "Target not found in list" for target 6

Here is my code. I cannot understand why 6 produces "Target not found in list" message when calling the last verify.

def linear_search(list, target): """ Returns the index position of the target if found, else returns None. """

for i in range(0, len(list)): if list[i] == target: return i return None

def verify(index): if index is not None: print("Target found at index: ", index) else: print("Target not found in list")

numbers = [1,2,3,4,5,6,7,8,9,10]

result = linear_search(numbers, 12) verify(result)

result = linear_search(numbers, 6) verify(result)

2 Answers

KRIS NIKOLAISEN
KRIS NIKOLAISEN
54,972 Points

The return statement exits immediately, so in linear_search anything other than the first list item (1) will return target not found. Think of return None as an else statement that will only execute once and halt the loop.

Brittney Coble
Brittney Coble
16,474 Points

Thank you! That helped.

Suhas Nukala
Suhas Nukala
2,786 Points
Suhas Nukala
Suhas Nukala
2,786 Points

Remove 'return None' in linear_search function.