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Input: malayalam Output: Yes Input: geeks Output: No
Method # 1
1) Find Backward Line
2) Check if the reverse and the original match. p>
Iterative Method: this method is provided by Ball Raza . Loop from start to length / 2 and check the first character to the last character of the string and the second to the second last, and so on .... If any character does not match, the string will not be a palindrome.
Below is the implementation of the above approach:
# function to check the line
# palindrome or not
# Loop from 0 to len / 2
# main function
Method using built-in function for call lines: this method is provided by Shariq Raza . In this method, the predefined function & # 39; & # 39; .join (reversed (string)) is used to reversed a string.
Below is the implementation of the above approach:
Method using one additional variable: In this method, the user takes a string character one by one and stores it in an empty variable. After saving all characters, the user compares both strings and checks if it is a palindrome or not.
This article is courtesy of Sahil Rajput . If you are as Python.Engineering and would like to contribute, you can also write an article using contribute.python.engineering or by posting an article contribute @ python.engineering. See my article appearing on the Python.Engineering homepage and help other geeks.
Please post comments if you find anything wrong or if you would like to share more information on the topic discussed above.
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Python program to check if a string is palindrome or not find: Questions
Finding the index of an item in a list
Given a list
["foo", "bar", "baz"] and an item in the list
"bar", how do I get its index (
1) in Python?
>>> ["foo", "bar", "baz"].index("bar") 1
Reference: Data Structures > More on Lists
Note that while this is perhaps the cleanest way to answer the question as asked,
index is a rather weak component of the
list API, and I can"t remember the last time I used it in anger. It"s been pointed out to me in the comments that because this answer is heavily referenced, it should be made more complete. Some caveats about
list.index follow. It is probably worth initially taking a look at the documentation for it:
list.index(x[, start[, end]])
Return zero-based index in the list of the first item whose value is equal to x. Raises a
ValueErrorif there is no such item.
The optional arguments start and end are interpreted as in the slice notation and are used to limit the search to a particular subsequence of the list. The returned index is computed relative to the beginning of the full sequence rather than the start argument.
Linear time-complexity in list length
index call checks every element of the list in order, until it finds a match. If your list is long, and you don"t know roughly where in the list it occurs, this search could become a bottleneck. In that case, you should consider a different data structure. Note that if you know roughly where to find the match, you can give
index a hint. For instance, in this snippet,
l.index(999_999, 999_990, 1_000_000) is roughly five orders of magnitude faster than straight
l.index(999_999), because the former only has to search 10 entries, while the latter searches a million:
>>> import timeit >>> timeit.timeit("l.index(999_999)", setup="l = list(range(0, 1_000_000))", number=1000) 9.356267921015387 >>> timeit.timeit("l.index(999_999, 999_990, 1_000_000)", setup="l = list(range(0, 1_000_000))", number=1000) 0.0004404920036904514
Only returns the index of the first match to its argument
A call to
index searches through the list in order until it finds a match, and stops there. If you expect to need indices of more matches, you should use a list comprehension, or generator expression.
>>> [1, 1].index(1) 0 >>> [i for i, e in enumerate([1, 2, 1]) if e == 1] [0, 2] >>> g = (i for i, e in enumerate([1, 2, 1]) if e == 1) >>> next(g) 0 >>> next(g) 2
Most places where I once would have used
index, I now use a list comprehension or generator expression because they"re more generalizable. So if you"re considering reaching for
index, take a look at these excellent Python features.
Throws if element not present in list
A call to
index results in a
ValueError if the item"s not present.
>>> [1, 1].index(2) Traceback (most recent call last): File "<stdin>", line 1, in <module> ValueError: 2 is not in list
If the item might not be present in the list, you should either
- Check for it first with
item in my_list(clean, readable approach), or
- Wrap the
indexcall in a
try/exceptblock which catches
ValueError(probably faster, at least when the list to search is long, and the item is usually present.)
One thing that is really helpful in learning Python is to use the interactive help function:
>>> help(["foo", "bar", "baz"]) Help on list object: class list(object) ... | | index(...) | L.index(value, [start, [stop]]) -> integer -- return first index of value |
which will often lead you to the method you are looking for.
The majority of answers explain how to find a single index, but their methods do not return multiple indexes if the item is in the list multiple times. Use
for i, j in enumerate(["foo", "bar", "baz"]): if j == "bar": print(i)
index() function only returns the first occurrence, while
enumerate() returns all occurrences.
As a list comprehension:
[i for i, j in enumerate(["foo", "bar", "baz"]) if j == "bar"]
Here"s also another small solution with
itertools.count() (which is pretty much the same approach as enumerate):
from itertools import izip as zip, count # izip for maximum efficiency [i for i, j in zip(count(), ["foo", "bar", "baz"]) if j == "bar"]
This is more efficient for larger lists than using
$ python -m timeit -s "from itertools import izip as zip, count" "[i for i, j in zip(count(), ["foo", "bar", "baz"]*500) if j == "bar"]" 10000 loops, best of 3: 174 usec per loop $ python -m timeit "[i for i, j in enumerate(["foo", "bar", "baz"]*500) if j == "bar"]" 10000 loops, best of 3: 196 usec per loop
Why is it string.join(list) instead of list.join(string)?
This has always confused me. It seems like this would be nicer:
my_list = ["Hello", "world"] print(my_list.join("-")) # Produce: "Hello-world"
my_list = ["Hello", "world"] print("-".join(my_list)) # Produce: "Hello-world"
Is there a specific reason it is like this?
It"s because any iterable can be joined (e.g, list, tuple, dict, set), but its contents and the "joiner" must be strings.
"_".join(["welcome", "to", "stack", "overflow"]) "_".join(("welcome", "to", "stack", "overflow"))
Using something other than strings will raise the following error:
TypeError: sequence item 0: expected str instance, int found
This was discussed in the String methods... finally thread in the Python-Dev achive, and was accepted by Guido. This thread began in Jun 1999, and
str.join was included in Python 1.6 which was released in Sep 2000 (and supported Unicode). Python 2.0 (supported
str methods including
join) was released in Oct 2000.
- There were four options proposed in this thread:
joinas a built-in function
- Guido wanted to support not only
tuples, but all sequences/iterables.
seq.reduce(str)is difficult for newcomers.
seq.join(str)introduces unexpected dependency from sequences to str/unicode.
join()as a built-in function would support only specific data types. So using a built-in namespace is not good. If
join()supports many datatypes, creating an optimized implementation would be difficult, if implemented using the
__add__method then it would ve
- The separator string (
sep) should not be omitted. Explicit is better than implicit.
Here are some additional thoughts (my own, and my friend"s):
- Unicode support was coming, but it was not final. At that time UTF-8 was the most likely about to replace UCS2/4. To calculate total buffer length of UTF-8 strings it needs to know character coding rule.
- At that time, Python had already decided on a common sequence interface rule where a user could create a sequence-like (iterable) class. But Python didn"t support extending built-in types until 2.2. At that time it was difficult to provide basic
iterableclass (which is mentioned in another comment).
Guido"s decision is recorded in a historical mail, deciding on
Funny, but it does seem right! Barry, go for it...
Guido van Rossum
join() method is in the string class, instead of the list class?
I agree it looks funny.
Historical note. When I first learned Python, I expected join to be a method of a list, which would take the delimiter as an argument. Lots of people feel the same way, and there‚Äôs a story behind the join method. Prior to Python 1.6, strings didn‚Äôt have all these useful methods. There was a separate string module which contained all the string functions; each function took a string as its first argument. The functions were deemed important enough to put onto the strings themselves, which made sense for functions like lower, upper, and split. But many hard-core Python programmers objected to the new join method, arguing that it should be a method of the list instead, or that it shouldn‚Äôt move at all but simply stay a part of the old string module (which still has lots of useful stuff in it). I use the new join method exclusively, but you will see code written either way, and if it really bothers you, you can use the old string.join function instead.
--- Mark Pilgrim, Dive into Python
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Tallinn | 2022-11-30
Thanks for explaining! I was stuck with Python program to check if a string is palindrome or not for some hours, finally got it done 🤗. Will use it in my bachelor thesis
Massachussetts | 2022-11-30
Maybe there are another answers? What Python program to check if a string is palindrome or not exactly means?. I am just not quite sure it is the best method
Rome | 2022-11-30
Simply put and clear. Thank you for sharing. Python program to check if a string is palindrome or not and other issues with sin was always my weak point 😁. Will use it in my bachelor thesis