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Given a collection of numbers, nums, that might contain duplicates, return all possible unique permutations in any order. Example 1: Input: nums = 1,1,2 Output: 1.
Question or problem about Python programming:
I know about itertools, but it seems it can only generate permutations without repetitions.
For example, I’d like to generate all possible dice rolls for 2 dice. So I need all permutations of size 2 of [1, 2, 3, 4, 5, 6] including repetitions: (1, 1), (1, 2), (2, 1)… etc
If possible I don’t want to implement this from scratch
How to solve the problem:
Solution 1:
You are looking for the Cartesian Product.
In mathematics, a Cartesian product (or product set) is the direct product of two sets.
In your case, this would be {1, 2, 3, 4, 5, 6}
x {1, 2, 3, 4, 5, 6}
.itertools
can help you there:
To get a random dice roll (in a totally inefficient way):
Solution 2:
You’re not looking for permutations – you want the Cartesian Product. For this use product from itertools:
Solution 3:
In python 2.7 and 3.1 there is a itertools.combinations_with_replacement
function:
Solution 4:
In this case, a list comprehension is not particularly needed.
Given
Code
Details
4 Permute 2
Unobviously, Cartesian product can generate subsets of permutations. However, it follows that:
- with replacement: produce all permutations nr via
product
- without replacement: filter from the latter
Permutations with replacement, nr
Permute 2.2 For Macos Catalina
Permutations without replacement, n!
Consequently, all combinatoric functions could be implemented from product
:
combinations_with_replacement
implemented fromproduct
combinations
implemented frompermutations
, which can be implemented withproduct
(see above)
Solution 5:
I think I found a solution using only lambdas
, map
and reduce
.
Essentially I’m mapping a first lambda function that given a row, iterates the columnns
then this is used as the output of a new lambda function
which is mapped across all the possible rows
Permute 2.2 For Macos Operating System
and then we reduce all the resulting lists into one.
Permute 3
even better
Can also use two different numbers.
Solution 6:
Permute App
First, you’ll want to turn the generator returned by itertools.permutations(list) into a list first. Then secondly, you can use set() to remove duplicates
Something like below: