n=input( )
k=list(n)
for j in range(len(k)):
if k[j]=="a":

k.pop(j)
print(k)
''' i want to print the list by removing a letter "a" but it is showing index error why ? help me out'''

You don't have to cast your input "n" to a list. Your input will be a string and in essence a string is just a list of characters.

So you could do the following:

n = input()

for char in n:

if char == "a":

# do your thing

@TipsyTapir that's a rather inconvenient way to do it. I would use string replace with a count of 1, if his question was worded correctly.

@highlight
print(input().replace("a", "", 1))

okay ... but why iam getting that error ..even the range is not crossing the index

@TipsyTapir While Python Strings are iterable and indexable, they are also immutable

@sharief
You modify the list you are iterating over. I will show you what happens (this is pseudocode, obviously):

n = "bar"
k=['b', 'a', 'r']
len(n) == 3

Steps of for j in range(3):
j = 0; k[0] = 'b'
j = 1; k[1] = 'a'; k.pop('a'); k = ['b', 'r']
j = 2; k[2] ⬅IndexError happens here, as k only has two elements, and not three.

What could you do instead?

Python Strings have a .replace method (e.g. n = n.replace("a", "").

Alternatively (if you don't actually have a string, but a list if something else) you could also do this:

while "a" in k:
k.pop(k.index("a"))

thank you 😀😀😀😀

@p100sch Yep should have read the complete question. I was more focusing on the index out of range error. Should have noticed he just wanted to replace some value. My bad 😜