 # Return 1 if a specific letter is in a string

<Below this line, add a link to the EXACT exercise that you are stuck at.>
I need a code that returns 1 when the number of z’s in the input string is greater than/equal to 3.
Otherwise, it shall return 0. You can find my code below.

<In what way does your code behave incorrectly? Include ALL error messages.>

<What do you expect to happen instead?>

If I , for instance, write punk(“zzzzz”) , it should return 1

``````
def punk(s):
count=0
lett=['z']
for char in s:
if char in lett:
if count==3:
print("number of z:" + str(count))
else:
return 0

``````

If you want to have a look at how `for` loops work (and following your code), you want to change a few things:

``````for char in s:
if count==3:
print("number of z:" + str(count))
``````

Firstly, you need to have something which actually increments `count`, otherwise it’ll remain at 0. So the first thing you should do after checking if the character is in `lett` is this: `count += 1`

Secondly, you should only have to check count once, and that should be after the `for` loop finishes (otherwise it’ll check count every time it cycles through a letter, which is a waste of memory).

Equally, your else statement should move along with it - you only want to return 0 once that for loop is finished.

Thirdly, you want to check if the `count` is greater than or equal to 3, not just equal to 3. So you want if `count >= 3`

Finally, if you want to return 1, simply place a `return 1` after your print statement.

So your code should look like this:

``````def punk(s):
count = 0
lett=['z']
for char in s:
if char in lett:
# increment count
if count >= 3:
# print statement followed by returning 1
else:
return 0
``````

There is of course another way you could do this. Python has a count function for strings, for example:

``````myString = "Hello World"
myString.count("l")
``````

In this case, `count()` would return 3. So you could use this in place of a `for` loop altogether.

Have a go and see if you can get your code working, good luck!