Help in XYZ problem in python?

Question

Edit: Without regex

Return True if the given string contains an appearance of "xyz" where the xyz is not directly preceeded by a period (.). So "xxyz" counts but "x.xyz" does not.

E.g:
xyz_there('abcxyz') → True
xyz_there('abc.xyz') → False
xyz_there('xyz.abc') → True

My mindless attempt:
It does't pass all conditions

https://code.sololearn.com/cu05NPBlYUql/?ref=app

Answer 1

Try this one:

https://code.sololearn.com/cAzxk02W6fUI/?ref=app

Answer 2

Regex!

https://code.sololearn.com/cErhYv2F92Ev/?ref=app

Answer 3

Russ, thanks a lot for your hint!!! Now it works.

Answer 4

Russ ohh i never thought of making it shorter. Nice!!

Answer 5

[Edited / corrected] - Here is my version:

test = ['.xyz.', 'abcxyz', 'abc.xyz', 'xyz.abc','.xyzabc', 'abc.xyzxyz', '.xyz.xyz', 'abc.xyzxyz.xyz']

for i in test:
print(f'{i} -> OK') if (i.count('xyz') - i.count('.xyz') > 0) else print(f'{i} -> NOT OK')
print()

https://code.sololearn.com/ca8NngqIpDeg/?ref=app

Answer 6

Bikash Rouniyar, You posted a question here in this thread. The better way would be if you post it as a separate thread. Thanks!

Answer 7

Russ Good one :) My attempt:
https://code.sololearn.com/c4iVsRm45zrl/?ref=app

Answer 8

PythonPip I suspect one of the failing tests might be from values like:

".xyzxyz"
"x.xyzxyz"
"xyz.xyz"
"x.xyz.xyz.xyzxyz"

Both ".xyz" and "xyz" without a preceeding "." exists. So the solution should also return True for these instances.

Answer 9

How about a little state machine FSM
States: -1,0,1,2
Start state 0
Accepted state 3
0->0 for each char except x and.
O->1 for x
0->-1 for .
1->2 for y
1->-1for .
1->0 for others
2->-1 for .
2->3 for z
2->0 for others
-1 -> -1 for .
-1 ->0 for others

Answer 10

Lothar I think change "== 1" to "> 0" and your code is good 👍

Answer 11

Cassandra If only '.x' is removed then the letters before '.x' will create problem if it's 'x'.

Consider this string: "x.xyz"

If ".x" would be removed then then string will read as "xyz", which will return True instead of False.

Answer 12

Kuba Siekierzyński It needs to catch "xyzabc" as true 😉

Answer 13

Russ Right, your way is more efficient 👍🏻

Answer 14

𝐊𝐢𝐢𝐛𝐨 𝐆𝐡𝐚𝐲𝐚𝐥 😅😅
It's an online website test. That won't allow this.

Answer 15

PythonPip I thought ".xyz" should return False?

Code Crasher Your code produces this:

djd.xyzxyz True
djdj.xyzjen True - should be False
jdjxyz.xyz True
djdjhexyz False - should be True
abcxyz False - should be True
xyzabc False - should be True

I think Valmob101 's solution is the cleanest.

def xyz_there(string):
return "xyz" in string.replace(".xyz", "")

Answer 16

Code Crasher the question says that, if there is xyz in string it should return True but if there is no xyz but there's .xyz then False.

means it should contain xyz to return True even if there are many .xyz

Did you understand or did i made you more confused?😅😅

Answer 17

Code Crasher If there is at least 1 instance of "xyz" appearing where it is not preceded by a ".", it should return True. Else, it should return False.

Answer 18

On the 'Discuss' tab, you should see a green circle at the bottom right. Hit that to ask a new question.

Answer 19

Moye Nkem Onyeka Not here, no. What you need to do is post the challenge to your own page, then post a link to it in this thread: https://www.sololearn.com/Discuss/1270852/?ref=app.

Answer 20

PythonPip Not "cleanly" I guess. But doable. Check updated code above.