I have one string 'abcdef'
Need to print element like first element, last element, second element, second last element
Expected out is afbecd
Another output for abcde Expected is aebdc
Can we do it in without creating a extra list
pseudo code:
str1 = 'abcdef'
i= 0
j = 1
new_str = ''
while (i < len(str1) && j > len(str1) and i!=j):
new_str = str1[i] + str1[j]
Without any spurious memory usage, you can use some lazy iterators/generators with reversed and zip
def interleave(s):
gen = (c for pair in zip(s, reversed(s)) for c in pair)
return "".join(next(gen) for _ in s)
>>> interleave("abcdef")
'afbecd'
>>> interleave("abcde")
'aebdc'
You can introduce more utils to shorten the code even more:
from itertools import chain
def interleave(s):
gen = chain.from_iterable(zip(s, reversed(s)))
return "".join(next(gen) for _ in s)
gen = cycle(f(s) for f in (iter, reversed))apply, then I think cycle(map(apply, (iter, reversed), repeat(s))) would work.With your approach you can do:
str1 = 'abcdef'
i= 0
j = len(str1) - 1
new_str = ''
while (j>i):
new_str += str1[i] + str1[j]
i+=1
j-=1
if len(str1) % 2 != 0:
new_str += str1[j]
print(new_str)
Output:
afbecd
if j == i: instead.The way I see it, you have two possibilities based on string length here but they really only differ in how you handle the middle character. Therefore, an input of abcdefg should give you an output of agbfced whereas the output for abcdef would be afbecd.
Given the above, I think this should work:
def zip_str(input_str):
width = len(input_str)
half_width = int(width / 2)
new_str = ""
for i in range(0, half_width):
new_str += input_str[i] + input_str[-1 - i]
return new_str if width % 2 == 0 else new_str + input_str[half_width]
A not particularly efficient but simple and imho fun way is to keep reversing the remaining string (Try it online!):
s = 'abcdef'
result = ''
while s:
*s, char = reversed(s)
result += char
print(result)
Or, since you actually say you want to print (Try it online!):
s = 'abcdef'
while s:
*s, char = reversed(s)
print(char, end='')