That removes all duplicate values, since they will hash to the same thing. This works by making a set() of the permutations. Using modules from itertools import permutations You should use generators instead of keeping everything in a list, this avoids high memory usage when working with large numbersįor perm in unique_perms(elements, unique): You could use a set() to have \$O(N)\$ lookup when finding items that are already seen. Use libraries when possible, you are currently reinventing the wheel of itertools.permutations Output_list, output_list_copy, temp_output =, , Start off with just the last element (c) in a set (), then add the second last element (b) to its front, end and every possible positions in the middle, making it and then in the same manner it will add the next element from the back (a) to each string in the set making it: There are many better solutions out there but I am interested in just the code review and how it can be made better. So the total steps in this invocation equal (IV) => The sequence is unchanged.Given a collection of numbers that might contain duplicates, returnįor example, have the following unique permutations: The recursive calls "generate(n - 1, A)" result in a right-rotation. The swaps that get executed in this invocation result in a right-rotation, see (III). The recursive calls "generate(n - 1, A)" leave the ordering unchanged (because it invokes generate with first argument being 1). Trivially, the ordering is unchanged after invoking the function. Now we can see why (I) and (II) are true: Taking the element at 0 out before the rotation, then after the rotation swapping it with the new element at 0 does not change the outcome (if rotating n times). If you have a sequence of length 5, then rotate it 5 times, it ends up unchanged. (IV) This series of steps leaves the sequence in the exact same ordering as before: Repeat n times: (III) This series of swaps result in a rotation to the right by one position: A Aįor example try it with sequence ABCD: A A: DBCA Just swap the elements at 0 and (n-1) in each iteration to produce a unique set of elements.įinally, let's see why the initial statements are true: Rotate-right So the element at index 0 will always be a different element automatically. The recursive call "generate(n - 1, A)" has rotated the elements right. So the for-loop can iteratively swap the element at i=0.(n-1) with the element at (n - 1) and will have called "generate(n - 1, A)" each time with another element missing. The recursive call "generate(n - 1, A)" does not change the order. (II) where n is even, rotates the elements to the right, for example ABCD becomes DABC. (I) where n is odd, leaves the elements in the exact same ordering when it is finished. Please just assume that these statements are true for a moment (i'll show that later): Each invocation of the "generate" function However, I think it is hard to understand it, so came up with an explanation that is hopefully easier to understand: I found an article that tries to explain it here: Why does Heap's algorithm work? Looking at the diagram on the Wikipedia page might help. You can prove that both of these facts are true using induction, although that doesn't provide any intuition as to why it's true. Whereas if the vector is of odd length, it will be simply swap the first and last elements: → Given a vector of even length, a full iteration of Heap's algorithm will rearrange the elements according to the rule → To see how Heap's algorithm works, you need to look at what a full iteration of the loop does to the vector, in both even and odd cases. See for the correct algorithm (at least, it's correct today) or see the discussion at Heap's algorithm permutation generator In either case, it does unnecessary work. Reverse the part of the permutation to the right of where that element was.īoth steps (1) and (3) are worst-case O(n), but it is easy to prove that the average time for those steps is O(1).Īn indication of how tricky Heap's algorithm is (in the details) is that your expression of it is slightly wrong because it does one extra swap the extra swap is a no-op if n is even, but significantly changes the order of permutations generated when n is odd. Swap that element with the smallest element to its right which is larger than it. Given some permutation, find the next one by:įinding the rightmost element which is smaller than the element to its right. The lexicographic order algorithm is extremely simple to describe. There is a much more intuitive algorithm which will produce permutations in lexicographical order although it is amortized O(1) (per permutation) instead of O(1), it is not noticeably slower in practice, and it is much easier to derive on the fly. Heap's algorithm is probably not the answer to any reasonable interview question.
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