![]() If statement's for loop repeats 'n' times to display chars from the example "ABC" i.e. So from point 2 to 4.2 total calls are 5 for each loop and total is 15 calls + main entry call = 16.Įach time loopCnt is 3 then if condition gets executed.įrom the diagram we can see loop count becoming 3 total 6 times i.e. Under each loop we are recursively calling with LpCnt + 1.Ĥ.1 When index is 1 then 2 recursive calls.Ĥ.2 When index is 2 then 1 recursive calls.In the else part in for loop we are repeating from 0 to 2 making 1 call each time.So calling with Index 0 and that is first call. Permutations method called from Main for first time.ResultString += "\nNo of Swap Calls : " + noOfSwapCalls ResultString += "\nNo of Char Display Calls : " + noOfCharDispla圜alls ResultString += "\nNo of For Loop Calls : " + noOfForLoopCalls ResultString += "\nNo of General Case Calls : " + noOfRecursiveCaseCalls ResultString += "\nNo of Base Case Calls : " + noOfBaseCaseCalls ResultString += "\nNo of Function Calls : " + noOfFunctionCalls ![]() string resultString = Permute("ABCDE".ToCharArray(), 0) string resultString = Permute("ABCD".ToCharArray(), 0) String resultString = Permute("ABC".ToCharArray(), 0) string resultString = Permute("AB".ToCharArray(), 0) string resultString = Permute("A".ToCharArray(), 0) Public static void StringPermutationsTest() Static void Swap(ref char Char1, ref char Char2) Permute(elementsList, (currentIndex + 1)) Swap(ref elementsList, ref elementsList) Static string Permute(char elementsList, int currentIndex)įor (int lpIndex = currentIndex lpIndex < elementsList.Length lpIndex++) I have written the sample in C# but easy to understand for most of the programmers. Also planning to explain the running time without focusing on Recursive Reconciliation. Though it is little old question and already answered thought of adding my inputs to help new visitors. Perhaps this visualization of the call graph with a slightly smaller string makes it more obvious how it works: For an explanation of the code I'll refer you to quasiverse's excellent answer. You have to run through the code by "hand" (using whatever tools you want - debuggers, paper, logging function calls and variables at certain points) until you understand it. We also need a for loop as we need to single out each character from the string.PaulR has the right suggestion. This can be done using the charAt function. We are taking a single character from the given string, starting with ‘a’, moving on to ‘c’ and finally visiting ‘e’. Translating these observations into a technical perspective Appending this to ‘e’ results in “ eac” and “ eca“. With “ac”, rearranging them results in “ac” and “ca”. If we single out the character ‘e’ in ace, we are left with “ac”. Appending this to ‘c’ results in “ cae” and “ cea“. With “ae”, rearranging them gives us “ae” and “ea”. If we single out the character ‘c’ in ace, we are left with “ae”. Finally appending each to “a” results in “ ace” and “aec“. ![]() If the input string was “aced”, we will get 24 permutations – 4 ! Observation about the permutationsĭid you notice that the first 2 permutations “ ace”, “ aec” have a as the first letter and the 2 other letters are then concatenated to the letter a.Įxtracting the first character ‘a’ from “ace” leaves us with the remaining characters “ce”. Note that the string “ace” is of length 3 and we get 6 different permutations of the same – 3 factorial. The string “ace” can be arranged as “ace”, “aec”, “cae”, “cea”, “eac”,”eca” – different arrangements of the characters a,c,e which make the string “ace”. Let’s take an example to understand this. Permutation of a string is arranging the characters of the string in different ways. What is a permutation ?ĭifferent ways of arranging a set of items is permutation. Along the way, I will also be explaining each line code and show you pictorial representations of the code execution so that you can visualize it better. Finally, we will write code in Java for the same. ![]() Like always, we will first start from the basics – Understand what is a permutation of a string, break the procedure down into smaller steps and understand the procedure of finding out one permutation. In this blog we are going to find out all the permutations of a given string. ![]()
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