![]() ![]() New-Object PSObject -Property = $stringBuilder. My PowerShell approach is pretty primitive, but works just fine for me. So I know that there are 6 possible permutations to “cat”. An easier way to look at it is like this: I use a factorial with the number of characters in the string to determine how many permutations there are. Knowing this, the formula that I needed consisted of this: The solution needs to handle the general case of an unknown number of elements with pipes. It is basically splitting on the commas and then if those elements have a pipe create permutations for every pipe delimited sub-element (of the remaining elements). On a whiteboard or piece of paper, write out the permutations for a string of length 3, like ‘abc’ (see above), and a string of length 4, like ‘abcd’. This means that each item in the string can only be used once, so if I chose “cat”, it can only have the “c”,”a” and “t” instead of possibly having all “c’s”, “a’s” and “t’s”. Given 12,34 need to get 4 strings: 1,3 1,4 2,3 2,4. In this case, I was aiming for a Permutation with no repetition. Number of Permutations With Repetitions In the aforementioned article, we derived a formula for calculating the number of distinct permutations of N elements as N, representing the product of all numbers from 1 to N, i.e., 123 (N-1)N. But first I had to figure out what the formula was to make this happen. We can see that there are only three distinct strings: ‘ulu’, ‘uul’, and ‘luu’. I really don’t need to calculate the number of permutations, but just for the heck of it, I through something quick together to get the possible combinations. Since my main goal was to take a given word (or string) and cycle through all of the possible combinations, I figure this would make for a fun attempt at a PowerShell function. all permutations are in lexicographic order.Somewhere a while back I was reading about working with trying to find all possible combinations of a given word, called Permutations. Print "The code is correct for |S| = %d." % nrElements # This line is only reached if no assertion failed, i.e. Result.append(p1) # add to list of permutationsįor i in xrange(len(result)-1): # check that permutations are in lexicographic order.Īssert result < result, "Permutations are not sorted, the code is incorrect."Īssert len(set(result)) = len(result), "Permutation contains multiple copies of an element, the code is incorrect."Īssert len(set(result)) = len(result), "Permutation contains multiple copies of an element, the code is incorrect." # check on last element P2 = kthpermrec(S, k) # compute k-th permutation recursivelyĪssert p1=p2 # make sure the recursive and non-recursive function yield the same permutation P1 = kthperm(S, k) # compute k-th permutation iteratively # This creates the k-th permutations for k=0.len(S)!, and then checks that the result is indeed in lexicographic order.įor k in xrange(factorial(nrElements)): # loop over all k=0.len(S)! For example, if you have an initial set of three stringsapple, banana, cherrythen there are a total of six permutations: Copy. A string permutation is a rearrangement of a set of strings. # compute the k-th permutation of S (all indices are zero-based)ĭef kthperm(S, k): # nonrecursive versionĭef kthpermrec(S, k): # recursive version The ability to programmatically create and use string permutations is an essential software testing skill. Here's a little Python code, implementing the algorithm above as well as its recursive version, then checking correctness for $\vert S\vert=10$ (this might take some time to run): The $k$-th permutation $P_k$ of a string $S$ can be computed as follows (assuming zero-based index):Įssentially, this finds the first element of the k-th permutation of S, and then recurses on the remaining string to find its first element.ĭepending on whether you start counting your permutations from 0 or 1, the answers is $(2, 7, 8, 3, 9, 1, 5, 6, 0, 4)$ or $(2, 7, 8, 3, 9, 1, 5, 6, 4, 0)$. We have a loop in the code that runs for n times as the length of the question string is n. So, the time complexity function for this printPermutations (str,asf) is denoted by T (n,0). Initially, the length of the question is n and the length of the asf string is 0. I think the above solutions are slightly off. In the code, we are having the function printPermutations (str,asf). Where did I go wrong in thinking or applying? However, according to the solution tester, (free reg. ![]() Which results in "2783905614" as my solution. I have an ordered set of symbols S =, which permutation of these symbols is at lexicographic position (1,000,000 - 2*9!) = 274,240?Ħ*8! = 241,920 is the largest multiple of 8! which is smaller than 274,240, so I need the 6th-smallest symbol of the remaining set, which is 7. ![]()
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