Valid Permutation of Parenthesis [duplicate]

Answer 1

A non-recursive solution in Python:

#! /usr/bin/python

def valid(state,N):
    cnt=0
    for i in xrange(N):
        if cnt<0:
            return False
        if (state&(1<

This basically examines the binary representation of each number in [0,2^n), treating a '1' as a '{' and a '0' as a '}' and then filters out only those that are properly balanced.