catalan

How can we generate all possibilities on braces ? N value has given to us and we have to generate all possibilities. Examples: 1) if N == 1, then only one possibility () . 2) if N==2, then possibilities are (()), ()() 3) if N==3, then possibilities are ((())), (())(),()()(), ()(()) ... Note: left and right braces should match. I mean )( is INVALID for the N==1 Can we solve this problem by using recurrence approach ? From wikipedia - A Dyck word is a string consisting of n X's and n Y's such that no initial segment of the string has more Y's than X's (see also Dyck language). For example, the

The problem is to find the n-th Catalan number mod m , where m is NOT prime , m = (10^14 + 7) . Here are the list of methods that I have tried: (max N = 10,000 ) Dynamic programming for table look-up, too slow Use Catalan formula ncr(2*n, n)/(n + 1) , again it wasn't fast enough due to the ncr function, can't speed up using exponentiation squaring because m is not prime. Hardcode a table of pre-generated Catalans , but it failed due to the file size limit. Recurrence relation C(i,k) = C(i-1,k-1) + C(i-1,k) , this is way too slow So I wonder is there any other faster algorithm to find the n-th

This came up while talking to a friend and I thought I'd ask here since it's an interesting problem and would like to see other people's solutions. The task is to write a function Brackets(int n) that prints all combinations of well-formed brackets from 1...n. For Brackets(3) the output would be () (()) ()() ((())) (()()) (())() ()(()) ()()() markt Took a crack at it.. C# also. public void Brackets(int n) { for (int i = 1; i <= n; i++) { Brackets("", 0, 0, i); } } private void Brackets(string output, int open, int close, int pairs) { if((open==pairs)&&(close==pairs)) { Console.WriteLine(output

For Binary trees: There's no need to consider tree node values, I am only interested in different tree topologies with 'N' nodes. For Binary Search Tree: We have to consider tree node values. I recommend this article by my colleague Nick Parlante (from back when he was still at Stanford). The count of structurally different binary trees (problem 12) has a simple recursive solution (which in closed form ends up being the Catalan formula which @codeka's answer already mentioned). I'm not sure how the number of structurally different binary search trees (BSTs for short) would differ from that of