Resolved: a different approach worked for me
I did a different approach, and I think it worked. However, I doubt whether it is linear or not. However, I think, based on the iteration, it can be considered linear.
found=[]
def linear_search(l,m):
for i in range(len(l)):
if l[i] !=m:
found.append(m)
else:
print(f'your requested item is the {i+1}th item of the list')
return m,i
Hey Arsalan,
Thank you for reaching out and sharing your approach.
The solution you've provided indeed solve the problem in a linear fashion--it goes through the list and checks whether the number we're on is the one we're searching for.
Here are a couple of question to think about that could make your code more efficient. I'll make use of the following list: [6,5,8,2,3,45,87,24,70]
1. If the number I'm searching for is 5 and I've found it's on position i=1, do I need to go through the rest of the list or I can terminate my search?
2. Do I need the found list? Do I need to append m to a list?
3. Do I need to return the variables m and i?
Kind regards,
365 Hristina
