Reverse a string without using reversed() or [::-1]?

Question

I came across a strange Codecademy exercise that required a function that would take a string as input and return it in reverse order. The only problem was you could not use the reversed method or the common answer here on stackoverflow, [::-1].

Obviously in the real world of programming, one would most likely go with the extended slice method, or even using the reversed function but perhaps there is some case where this would not work?

I present a solution below in Q&A style, in case it is helpful for people in the future.

Solution 1

You can also do it with recursion:

def reverse(text):
    if len(text) <= 1:
        return text

    return reverse(text[1:]) + text[0]

And a simple example for the string hello:

   reverse(hello)
 = reverse(ello) + h           # The recursive step
 = reverse(llo) + e + h
 = reverse(lo) + l + e + h
 = reverse(o) + l + l + e + h  # Base case
 = o + l + l + e + h
 = olleh

Solution 2

Just another option:

from collections import deque
def reverse(iterable):
    d = deque()
    d.extendleft(iterable)
    return ''.join(d)

Solution 3

Use reversed range:

def reverse(strs):
    for i in xrange(len(strs)-1, -1, -1):
        yield strs[i]
...         
>>> ''.join(reverse('hello'))
'olleh'

xrange or range with -1 step would return items in reversed order, so we need to iterate from len(string)-1 to -1(exclusive) and fetch items from the string one by one.

>>> list(xrange(len(strs) -1, -1 , -1))
[4, 3, 2, 1, 0]  #iterate over these indexes and fetch the items from the string

One-liner:

def reverse(strs):
    return ''.join([strs[i] for i in xrange(len(strs)-1, -1, -1)])
... 
>>> reverse('hello')
'olleh'

Solution 4

EDIT

Recent activity on this question caused me to look back and change my solution to a quick one-liner using a generator:

rev = ''.join([text[len(text) - count] for count in xrange(1,len(text)+1)])

Although obviously there are some better answers here like a negative step in the range or xrange function. The following is my original solution:

---

Here is my solution, I'll explain it step by step

def reverse(text):

    lst = []
    count = 1

    for i in range(0,len(text)):

        lst.append(text[len(text)-count])
        count += 1

    lst = ''.join(lst)
    return lst

print reverse('hello')

First, we have to pass a parameter to the function, in this case text.

Next, I set an empty list, named lst to use later. (I actually didn't know I'd need the list until I got to the for loop, you'll see why it's necessary in a second.)

The count variable will make sense once I get into the for loop

So let's take a look at a basic version of what we are trying to accomplish:

It makes sense that appending the last character to the list would start the reverse order. For example:

>>lst = []
>>word = 'foo'
>>lst.append(word[2])
>>print lst
['o']

But in order to continue reversing the order, we need to then append word[1] and then word[0]:

>>lst.append(word[2])
>>lst.append(word[1])
>>lst.append(word[0])
>>print lst
['o','o','f']

This is great, we now have a list that has our original word in reverse order and it can be converted back into a string by using .join(). But there's a problem. This works for the word foo, it even works for any word that has a length of 3 characters. But what about a word with 5 characters? Or 10 characters? Now it won't work. What if there was a way we could dynamically change the index we append so that any word will be returned in reverse order?

Enter for loop.

for i in range(0,len(text)):

    lst.append(text[len(text)-count])
    count += 1

First off, it is necessary to use in range() rather than just in, because we need to iterate through the characters in the word, but we also need to pull the index value of the word so that we change the order.

The first part of the body of our for loop should look familiar. Its very similar to

>>lst.append(word[..index..])

In fact, the base concept of it is exactly the same:

>>lst.append(text[..index..])

So what's all the stuff in the middle doing?

Well, we need to first append the index of the last letter to our list, which is the length of the word, text, -1. From now on we'll refer to it as l(t) -1

>>lst.append(text[len(text)-1])

That alone will always get the last letter of our word, and append it to lst, regardless of the length of the word. But now that we have the last letter, which is l(t) - 1, we need the second to last letter, which is l(t) - 2, and so on, until there are no more characters to append to the list. Remember our count variable from above? That will come in handy. By using a for loop, we can increment the value of count by 1 through each iteration, so that the value we subtract by increases, until the for loop has iterated through the entire word:

>>for i in range(0,len(text)):
..        
..      lst.append(text[len(text)-count])
..      count += 1

Now that we have the heart of our function, let's look at what we have so far:

def reverse(text):

    lst = []
    count = 1

    for i in range(0,len(text)):

        lst.append(text[len(text)-count])
        count += 1

We're almost done! Right now, if we were to call our function with the word 'hello', we would get a list that looks like:

['o','l','l','e','h']

We don't want a list, we want a string. We can use .join for that:

def reverse(text):

    lst = []
    count = 1

    for i in range(0,len(text)):

        lst.append(text[len(text)-count])
        count += 1

    lst = ''.join(lst) # join the letters together without a space
    return lst

And that's it. If we call the word 'hello' on reverse(), we'd get this:

>>print reverse('hello')
olleh

Obviously, this is way more code than is necessary in a real life situation. Using the reversed function or extended slice would be the optimal way to accomplish this task, but maybe there is some instance when it would not work, and you would need this. Either way, I figured I'd share it for anyone who would be interested.

If you guys have any other ideas, I'd love to hear them!

Solution 5

Only been coding Python for a few days, but I feel like this was a fairly clean solution. Create an empty list, loop through each letter in the string and append it to the front of the list, return the joined list as a string.

def reverse(text):
backwardstext = []
for letter in text:
    backwardstext.insert(0, letter)
return ''.join(backwardstext)

View more solutions

Comments

[samrap]

can you explain the xrange part? I saw a similar answer in another post but didn't understand what was going on

[Blender]

Recursive golf: f=lambda t:len(t)>1and f(t[1:])+t[0]or t

[Burhan Khalid]

You don't need that inner list.

[Ashwini Chaudhary]

@BurhanKhalid For efficiency purpose we do.

[Burhan Khalid]

This is a case of premature optimization.

[samrap]

wow this is the coolest answer on here, very unique. One question though, why is it join item[1] and not just join item?

[Brian Peterson]

Best answer by far. Python wasn't geared for this kind of low-level programming without the use of its helpful constructs, but this is by far the best way to do so if you must.

[CT Zhu]

The items will be like (0,'a'), tuples of a int (the index) and a str (the value), so you can't just .join() them. :-P

[Ketan Maheshwari]

I think this line is missing a -1 in the subscript of text: lst.append(text[len(text)-count])

[samrap]

No, that would only be necessary if count started at 0, it starts at 1

[samrap]

Looking back, this can be reduced to one line: [text[len(text) - count] for count in xrange(1,len(text)+1)]

[Ketan Maheshwari]

True. I find this slightly changed version more readable somehow: [text[len(text) - 1 - count] for count in xrange(len(text))]

[samrap]

Definitely more readable, though now that I have more Python experience I would go with a more readable comprehension like some of the answers here

[samrap]

That's great! Keep up the work, Codecademy was a great resource for me

[samrap]

That's a good one, one way to save the string more elegantly would be to turn your loop into a list comprehension and then return a str of the list with the join function: return ''.join([x[n] for n in range(len(x)-1, -1, -1)]) This is a simple one-liner. Also there is no need for str() in your range assuming x is already a string. Not sure if you've done list comprehensions yet in Python. When you have this will make more sense. Great job! :)

[Fadeiar]

Thank's a lot for your reply i have just moved on with the challenges on codeacademy and I could really make use of your hint :). After reading a little bit on list comprehension I was able to use your input for a function that kicks out all vowels in a string. def anti_vowel(text): vowels = "aeiou" return "".join(n for n in text if not n.lower() in vowels) Thanks a lot @samrap (can't give reputation points yet)

[samrap]

nice! I'm so stoked that people continue to find this a useful and interesting thread. Love seeing all the answers!

[Emanuele Paolini]

I think this solution requires O(n^2) memory... this could be very bad for long strings.

[Deepend]

It is better for the OP if you can add some more information on an Answer such as this. Others looking at this in future will all benefit.

[Blender]

@EmanuelePaolini: [::-1] is the only thing you should be using. All of these answers are just different ways of reversing a string, there are no speed guarantees.

[samrap]

Agreed with @deepend also I like this answer it is similar to the way it is done in K&R's C Programming Language book

[samrap]

+1 Didn't even know you could do this, or maybe I did at one point I just haven't worked with Python in 4 or 5 months. Either way, good job

[samrap]

Nice one! Liking the creativity on this one

[AdjunctProfessorFalcon]

@Blender Thanks for this breakdown, appreciate it. Just to be clear on what's happening here: is the text variable for text[0] basically storing each character it slices off as the recursive function calls keeps cycling through the string? or is it that with each recursive call another text[0] is being added?

[Blender]

@ChrisPurrone: The recursive function doesn't really "cycle" through the string (in CPython). Each time you call a function, a stack frame is created (which contains the local variables and the return value of your function). Your function is recursive, so each time you return something a new stack frame is created, but since the return value is dependent on another function's return value, the old stack frames still need to stick around. That's why I wrote the long chain of + h's. They aren't really concatenated until the very end, which is when the final function call returns a string.

[AdjunctProfessorFalcon]

@Blender So how you have all the strings being concatenated (until the final function call) is really just an way of describing how the strings that are sliced off by the text[0] are being stored in memory in each stack frame? Why is it that all the strings are only concatenated at the very end instead each string (in each stack frame) being added together? Sorry, newbie so struggling to understand how reverse(text[1:]0 and text[0] are working together here.

[Blender]

@ChrisPurrone: CPython doen't do any sort of recursion optimization at all (from what I remember). In fact, CPython doesn't even optimize tail recursive functions, which can be translated easily into equivalent iterative functions.

[krethika]

For my solution, I refactored this into a while loop to avoid recursion error on long strings.

[Pang]

The question says "without using reversed or [::-1]".

[simple noobie]

My original query was how to reverse a string and this is where I was led.

[Amit]

One of the fastest.

[Amit]

Simplest and one of the fastest!

[SamTheProgrammer]

Don't use str for a variable since it is already part of the Python keywords list. Try using new_string