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.

Answer 1

Try this simple and elegant code.

my_string= "sentence"
new_str = ""
for i in my_string:
    new_str = i + new_str
print(new_str)

Answer 2

I have got an example to reverse strings in python from thecrazyprogrammer.com:

string1 =  "IkNoWwHeReYoUlIvE"
string2 = ""

i = len(string1)-1

while(i>=0):
  string2 = string2 + string1[i]
  i = i-1

print("original = " + string1)
print("reverse  = " + string2)

Answer 3

simply run this but it would print each character in a separate line the second version prints it in one line.

def rev(str):
        for i in range(0,len(str)):
            print(list(str)[len(str)-i-1])

Print in one line:

def rev(str):
    rev = list()
    for i in range(0,len(str)):
        rev.append((list(str)[len(str)-i-1]))
    print(''.join(rev))

Answer 4

Simple Method,

>>> a = "hi hello"
>>> ''.join([a[len(a)-i-1] for i,j in enumerate(a)])
'olleh ih'
>>> 

Answer 5

You can simply use the pop as suggested. Here are one liners fot that

chaine = 'qwertyuiop'
''.join([chaine[-(x + 1)] for x in range(len(chaine))])
'poiuytrewq'
gg = list(chaine)
''.join([gg.pop() for _ in range(len(gg))])
'poiuytrewq'

Answer 6

I prefer this as the best way of reversing a string using a for loop.

def reverse_a_string(str): 

    result=" "
    for i in range(len(str),1,-1):
        result= result+ str[i-1]
    return result

print reverse_a_string(input())

Answer 7

Here is one using a list as a stack:

def reverse(s):
  rev = [_t for _t in s]
  t = ''
  while len(rev) != 0:
    t+=rev.pop()
  return t

Answer 8

We can first convert string into a list and then reverse the string after that use ".join" method to again convert list to single string.

text="Youknowwhoiam"
lst=list(text)
lst.reverse()
print("".join(lst))

It will first convert "youknowwhoiam" into ['Y', 'o', 'u', 'k', 'n', 'o', 'w', 'w', 'h', 'o', 'i', 'a', 'm']

After it will reverse the list into ['m', 'a', 'i', 'o', 'h', 'w', 'w', 'o', 'n', 'k', 'u', 'o', 'Y']

In last it will convert list into single word "maiohwwonkuoY "

For more explanation in brief just run this code:-

text="Youknowwhoiam"
lst=list(text)
print (lst)
lst.reverse()
print(lst)
print("".join(lst))

Answer 9

__author__ = 'Sreedhar'
#find reverse of the string with out using index or reverse function?

def reverse(text):
    revs = ""
    for i in text:
        revs = i + revs
    return revs

strig = raw_input("enter anything:")

print reverse(strig)

#index method
print strig[::-1]

Answer 10

enterString=input("Enter a string to be reversed:")
string_List=list(enterString)
revedString=[]

for i in range(len(string_List)):
    revedString.append(string_List.pop())
#Pop last element and append it to the empty list
print "reversed string is:",''.join(revedString)
#as the out come is in list convert it into string using .join

Sample Output:
>>>Enter a string to be reversed :sampleReverse1  
>>>reversed string is:1esreveRelpmas 

Answer 11

Easiest way to reverse a string:

    backwards = input("Enter string to reverse: ")
    print(backwards[::-1])

Answer 12

m = input("Enter string::")
def rev(s):
    z = -1
    x = len(s)
    while x != 0:
        print(s[z], end='')
        z = z-1
        x = x-1
rev(m)

Answer 13

You can do simply like this

def rev(str):
   rev = ""
   for i in range(0,len(str)):
   rev = rev + str[(len(str)-1)-i]
   return rev

Answer 14

Blender's answer is lovely, but for a very long string, it will result in a whopping RuntimeError: maximum recursion depth exceeded. One might refactor the same code into a while loop, as one frequently must do with recursion in python. Obviously still bad due to time and memory issues, but at least will not error.

def reverse(text):
    answer = ""
    while text:
        answer = text[0] + answer
        text = text[1:]
    return answer

Answer 15

This is the simplest way in python 2.7 syntax

def reverse(text):

      rev = ""
      final = ""

      for a in range(0,len(text)):
            rev = text[len(text)-a-1]
            final = final + rev
return final

Answer 16

reduce(lambda x, y : y + x, "hello world")

Answer 17

def reverse(s):
    return "".join(s[i] for i in range(len(s)-1, -1, -1))

Answer 18

Here's my contribution:

def rev(test):  
    test = list(test)
    i = len(test)-1
    result = []

    print test
    while i >= 0:
        result.append(test.pop(i))
        i -= 1
    return "".join(result)

Answer 19

This is my solution using the for i in range loop:

def reverse(string):
    tmp = ""
    for i in range(1,len(string)+1):
        tmp += string[len(string)-i]            
    return tmp

It's pretty easy to understand. I start from 1 to avoid index out of bound.

Answer 20

def reverseThatString(theString):
    reversedString = ""
    lenOfString = len(theString)
    for i,j in enumerate(theString):
        lenOfString -= 1
        reversedString += theString[lenOfString]
    return reversedString

Answer 21

My solution:

s = raw_input("Enter string ")
print
def reverse(text):

st = ""  
rev = ""  
count = len(text)  
print "Lenght of text: ", len(text)  
print  
for c in range(len(text)):  
    count = count - 1  
    st = st + "".join(text[c])  
    rev = rev + "".join(text[count])  
    print "count:       ", count  
    print "print c:     ", c  
    print "text[c]:     ", text[c]  
    print  
print "Original:    ", st  
print "Reversed:    ", rev  
return rev  

reverse(s)

Result screen

Enter string joca

Lenght of text: 4

count: 3
print c: 0
text[c]: j

count: 2
print c: 1
text[c]: o

count: 1
print c: 2
text[c]: c

count: 0
print c: 3
text[c]: a

Original: joca
Reversed: acoj
None

Answer 22

The way I can think of without using any built-in functions:

a = 'word'
count = 0
for letter in a:
    count += 1

b = ''
for letter in a:
    b += a[count-1]
    count -= 1

And if you print b:

print b
drow

Answer 23

You've received a lot of alternative answers, but just to add another simple solution -- the first thing that came to mind something like this:

def reverse(text):
    reversed_text = ""   

    for n in range(len(text)):
        reversed_text += text[-1 - n]

    return reversed_text

It's not as fast as some of the other options people have mentioned(or built in methods), but easy to follow as we're simply using the length of the text string to concatenate one character at a time by slicing from the end toward the front.

Answer 24

I used this:

def reverse(text):
s=""
l=len(text)
for i in range(l):
    s+=text[l-1-i]
return s

Answer 25

Pointfree:

from functools import partial
from operator import add

flip = lambda f: lambda x, y: f(y, x)
rev = partial(reduce, flip(add))

Test:

>>> rev('hello')
'olleh'

Answer 26

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)

Answer 27

You can simply reverse iterate your string starting from the last character. With python you can use list comprehension to construct the list of characters in reverse order and then join them to get the reversed string in a one-liner:

def reverse(s):
  return "".join([s[-i-1] for i in xrange(len(s))])

if you are not allowed to even use negative indexing you should replace s[-i-1] with s[len(s)-i-1]

Answer 28

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

Answer 29

Not very clever, but tricky solution

def reverse(t):
    for j in range(len(t) // 2):
        t = t[:j] + t[- j - 1] + t[j + 1:- j - 1] + t[j] + t[len(t) - j:]
    return t

Answer 30

All I did to achieve a reverse string is use the xrange function with the length of the string in a for loop and step back per the following:

myString = "ABC"

for index in xrange(len(myString),-1):
    print index

My output is "CBA"