![]() That's about 54 microseconds, about 3.7 million times faster than the code in the post. Third, counting them directly: > timeit(lambda:permutation_count('AAABBCDEF'), number=1) Second, the revised code using a set: > timeit(lambda:len(set(permutations('AAABBCDEF'))), number=1) > timeit(lambda:perms('AAABBCDEF'), number=1) First, your code from the post: > from timeit import timeit """Return the number of different permutations of s."""Īnd since it doesn't have to construct all the permutations, it can count large numbers of permutations in a fraction of a second: > permutation_count('AAAABBBCCDDEEFFGHIJKLMNOPQRSTUVWXYZ')Ī quick timing comparison. In that case, your output must be a Latin rectangle or a Latin square.These are easy to generate: start by constructing a Latin square, shuffle the rows, shuffle the columns, and then keep just the first r rows. So that's easy to program, using collections.Counter to count the number of occurrences of each character in the string, and math.factorial to compute the factorials: from collections import Counter Lets starting by solving the case of generating n or fewer rows first. ![]() So if we go back to the situation where A and a are indistinguishable, there are \$. Two is of course the number of permutations of two items: \$2! = 2\$. You'll see that these permutation come in twos: one with A before a (I've put these in the top row) and the other with a before A (in the bottom row). Let's distinguish the two copies of A (for the moment) by writing one of them as a and generate all \$4! = 4 × 3 × 2 × 1 = 24\$ permutations: AaBC AaCB ABaC ABCa ACaB ACBa BAaC BACa BCAa CAaB CABa CBAaĪABC aACB aBAC aBCA aCAB aCBA BaAC BaCA BCaA CaAB CaBA CBaA Suppose we are given the string AABC, with a repeated letter. In the actual question, the string might have repetitions. We call this "3 factorial" and write \$3! = 6\$. Use this: import itertools for x in itertools.permutations('1234'): print (''. join() DOCS, itertools.permutations DOCS. Use join() that return a string which is the concatenation of the strings in the iterable iterable. Then the permutations are: ABC ACB BAC BCA CAB CBAīut we can count them by observing that there are 3 choices for the first letter, 2 for the second, and one for the third, hence the total number of permutations is \$3 × 2 × 1 = 6\$. itertools.permutations takes an iterable and returns an iterator yielding tuples. ![]() Supposing that you knew that the string you are given has no repeated characters, for example ABC. Let's take a simpler example to show how this might work. It would be better to count them directly, via some kind of a mathematical formula. But your implementation goes and constructs all the permutations themselves, only to throw them all away again. Public Discord community with over 25,000. In this question you are asked to compute the number of permutations. Code solutions for 14 languges, including Python, Java, JavaScript and C++. One of the best ways to make a program faster is not to compute things that you don't have to.
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