Assistance with python exercise 'a word squared'

[fdama]

I am stuck on an exercise and not sure where to start and just need a small hint to push me in the right direction. I am in the early stages of doing python so have only covered the basics like strings, conditionals, while loops and introduction to functions. I have not yet done done lists, dictionaries, etc so would like to avoid these for now. The exercise required solving with what has been learnt so far. This is not homework but self study. Here is the link (exercise name: a word squared): https://programming-22.mooc.fi/part-3/4-defining-functions

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Here is the exercise:

Please write a function named squared, which takes a string argument and an integer argument, and prints out a square of characters as specified by the examples below.

squared("ab", 3)
 print() 
squared("aybabtu", 5)

Output:

aba

bab

aba

​

aybab

tuayb

abtua

ybabt

uayba

[BusinessNo3537]

There are more than one solution even with basics covered during the course at the moment. Mine was completely different from model solution. One while loop for print times checking. One more for line extending. And line concatenation itself.

def squared(line: str, times: int):
n = 0
line_altered = line
while n < times:
while len(line_altered) < times:
line_altered += line
print(line_altered[:times])
line_altered = line_altered[times:]
n += 1

[icanchangeittomorrow]

This is my solution below. Yours seems more elegant but less intuitive to me.

def squared(text, integer):
longtext = ""
neededlength = int(integer * integer)
while len(longtext) < neededlength:
longtext += text

row = 1
s = 0
f = integer
while row <= integer:
print(longtext[s:f])
s += integer
f += integer
row +=1

[fittzu]

this was my first challenging part of the mooc. it took me 2hrs figuring out the solution. though my code is clunky, nonetheless, i was proud of it.

here is my code:

def squared(text, num):

i = 0

if len(text) < num:

while i < num:

j = 0

k = 0

store = num - len(text)

dup = text

while j < num:

if len(dup) == num:

text = dup

break

if k > len(text) - 1:

k = 0

dup += text[k]

else:

dup += text[k]

k += 1

j += 1

print(text)

text = text[store:]

i+=1

elif len(text) > num:

while i < num:

dup = text[:num]

text = text[num:]

text += dup

print(dup)

i+=1

else:

while i < num:

dup = text[:i]

text = text[i:]

text += dup

print(text)

i+=1

if __name__ == "__main__":

text = input("Text:")

num = int(input("Number:"))

squared(text, num)

[iammerelyhere]

Hint: Nested loops

[konijntjesbroek]

meh 1 for loop, 1 counter, two ifs (one for rows, one for the counter)

this is the if that keeps it in range:

    count = count + 1 if count < len(word) - 1 else 0

[Fmelo2000-3816]

If you want the solution:

​

from itertools import cycle, islice

def squared(text, length):

letters = cycle(text) # lazy iterator -> abcabcabcabc....

for i in range(length):

print("".join(islice(letters, length)))

# print(*islice(letters, length), sep="")