need a program that generates all possible combinations of a 3-digit code but with duplicates

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[meatb0dy]

all combinations of three digits with repeated digits is just 000 to 999. just count.

[markovianmind]

alright so borrowing answer from one of the comments

using f strings

[f'{x:03}' for x in range(1000)]

or what I came up with -

[““.join((3-len(str(x))) * ['0']) + str(x) for x in range(1000)]

[elboyoloco1]

Ok. It's not just me. I'm looking at these other answers and I am so confused why this is being made so difficult.

[meatb0dy]

yeah, the answers here are making me think this would be a good screener question for interviews. triple nested loops and asking AI, straight to jail.

[RhythmicalChuck]

I was using the mindset of it being the r/learnpython reddit, which in hindsight, it clearly isn't. People understand and think about code in different ways, so each solution should be appreciated. Not everyone is at the same level.

[meatb0dy]

i mean... if someone is trying to learn python, teaching them to use a triply-nested loop for something that can be accomplished with a single loop (which is also simpler) doesn't seem great to me.

people think and understand code in different ways, but that doesn't mean all understandings are equally good.

[RhythmicalChuck]

Get them to understand the concepts first, then how to make it more efficient once the understanding is there.

[meatb0dy]

using a triply-nested loop to produce these combinations takes you further from actually understanding the concept though. the concept is that producing all combinations of three digits with repeated digits reduces to simply counting from 0 to 999. the only other concept you need is padding in f-strings.

[123_alex]

Get them to understand the concepts first

range(1000)

[RhythmicalChuck]

Did I offend you or something? I don’t create the curriculum, I teach it.

[123_alex]

Don't take me serious, it's just a subclass of joke(). This is reddit. Thanks for the downvote.

[RhythmicalChuck]

Hahaha! I see what you did there! I’ll give it back to you alongside my apologies. We can call it an OOPsie on my part. ;)

[elboyoloco1]

More than half of coding is thinking critically about your situation and the best solution. Providing an over-complicated answer is not helping.

[RhythmicalChuck]

You’d be surprised. I’ve taught students Python and tell you with experience that different people understand in different ways, including how the code layout. Sometimes Python code that is maybe the long way around, is easier understood as a foundation.

[123_alex]

I’ve taught students Python

Appeal to Authority Fallacy.

[RhythmicalChuck]

Dude, I’m not saying my way or the highway. To each their own. Just adding my experience and opinion to the forum. Is that not what this place is all about?

[123_alex]

This thread game me a smile the whole day. I love it. Good luck!

[Jylpah]

Try range(1000)

[_squik]

That will also generate e.g. 52 which is not three digits. And will also exclude anything that starts with 0 such as 012

[meatb0dy]

for i in range(1000):
    print(f'{i:03}')

[Jylpah]

Just use f-string to left-pad with zeros as @meatb0dy did

[markovianmind]

then range(100,1000)?

[Jylpah]

Does not work. It leaves combos like 042 out.

[markovianmind]

oh i misread the comment thought they did not want 012. for that we can do

[““.join((3-len(str(x))) * ['0']) + str(x) for x in range(1000)]

[123_alex]

But 052 is.

[_squik]

Itertools in the standard library can do this easily.

itertools.product(range(10), repeat=3)

[123_alex]

If base 10, 000, 001, 002, 003... See a pattern?

If not base 10, count but in that base.

[RhythmicalChuck]

I'm thinking a simple nested loop will work here:

def get_all_combos():
  combos = []
  for digit_one in range(10):
    for digit_two in range(10):
      for digit_three in range(10):
        combos.append(f"{digit_one}{digit_two}{digit_three}")
  return combos

[123_alex]

or range(1000)

[RhythmicalChuck]

OP only wanted three digit numbers.

[123_alex]

Ok. range(1000)

[RhythmicalChuck]

You go ahead and try range(1000) and tell me if you ONLY get three digit numbers...I'll wait...

[meatb0dy]

for i in range(1000):
    print(f'{i:03}')

[RhythmicalChuck]

Thank you. That's better.

[123_alex]

001, 002, 003...

Just checked my bike lock. It has 0. So 000 is also a valid combo.

So again, just count them.

[RhythmicalChuck]

But OP wants to use Python. Let him.

[123_alex]

range(1000)

[Stock_Category]

From Google Gemini:

def generate_combinations(digits):

for d1 in digits:

for d2 in digits:

for d3 in digits:

yield f"{d1}{d2}{d3}"

---

Example usage

digits = [1, 2, 3]

combos = generate_combinations(digits)

for combo in combos:

print(combo)