Stack A has the entries a,b,c ( with a on the TOp) Stack B is empty. an entry pooped out of the stack A can be printed immediatly or pushed to stack B. An Entry popped out of the stack B can only be printed. In this Arrangement ,if Stack A has 4 entries, then number of possible permutation will be
(a) 24 (b) 12 (c) 21 (d) 14 for 3 entries ..solution is 5.....abc,cba,bac,bca,acb....(except cab)..its like 3! - 1 (n! - 1 for n=3) ..is there any easy way to find for 4 entries ???...also what is the general solution ?? -- You received this message because you are subscribed to the Google Groups "Algorithm Geeks" group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
