![]() We have "A" and "N" repeated, meaning it adds more permutations to the outcome. We have $4$ characters so since we have $4$ options for the first character, $3$ for the second, $2$ for the third and $1$ for the last we have $4!$ different permutations.īut some of the characters are duplicates. If we asssume the string "ANNA" and we want the count of the permutation of duplicate items. Where do the factorials in the denominator exactly come from? Otherwise, the recursion is simple here running n deep and each level running a loop for n, n-1, n-2, n-3.2,1 times.I am not really sure I fully understand the formula for finding the number of permutation of duplicate items. With respect to string manipulation and permutation, think of the string as simply a 'set' of chars. this window gets decreasing at each level. At each level, you are running a piece of code, here you are running a for loop n-i times at each level. Think about the recursion as simply a number of levels. I have given some static CallCnt variables and the table to understand each line execution in detail.Įxperts, feel free to edit my answer or comment if any of my details are not clear or incorrect, I am happy correct them.ĭownload the sample code and other samples from here Total 6 times (Factorial times) we enter into if to display the permutations. 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.
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