how can I make a ab aba ... numbers?

Question

hi,
I want to make numbers with a & b so that for exp:
n=3
aba
n=4
abab
n=2
ab
n=5
ababa
without using for & while and str😊

Answer 1

Flandre Scarlet I don't think the problem was about that. But even if it were, reduce would do it just fine

reduce(lambda x, k: 10*x+b if k%2 else 10*x+a, range(n), 0)

Answer 2

Kishalaya Saha i think the question is like this:

supposed a=1, b=2, k=3, we should get 121, and not allowed to use int("121")🤔

Answer 3

Kishalaya Saha seems it's exactly what he want😂

Answer 4

I would do it with recursion🤔

aba = lambda a,b,k: k and [b,a][k%2] + 10*aba(a,b,k-1)
https://code.sololearn.com/c0Pkz7a6q49V

Answer 5

Or,
"ab" * (n//2) + "a" * (n%2)

Answer 6

The numbers are not converted to strings. The quotation marks are around ab (and also a in my case). The numbers remain numbers. "ab" * (n//2) concatenates the string "ab" with itself n//2 times. If n//2 were treated as a string, that wouldn't even make sense!

Regarding reduce, this should work:

from functools import reduce

n = 5
print(reduce(lambda x, k: x+"b" if k%2 else x+"a", range(n), ""))

But obviously it still uses strings, and is not better in any way than Bennet's method.

Answer 7

thanks a lot ... I was searching for reduce solution 😊

Answer 8

thanks but I forget that I dont have permission to use str

Answer 9

But we're not using the str() function!

Answer 10

I wrote a code for it but it is realy prolix, can you solve it by using reduce func?

Answer 11

we are getting numbers as string by using " "

Answer 12

thanks for explaining this 👌👌

Answer 13

You're welcome, shima 😊

Answer 14

Flandre Scarlet , oh, okay then! 😂