A permutation p1
of 1,2,...,n is even
) if it can be returned to the original order 1,2,...,n using an even (resp. odd) number of interchanges of pairs of elements. It is known that every permutation is either even or odd (and therefore not both).
The sign, or signum (for those who want to be fancy), of a permutation is +1 for an even permutation and -1 for an odd permutation.
You are given the integer n and the array p initialized to the integer values p=p1, p=p2, ..., p[n]=pn which are guaranteed to be a permutation of 1,2,...,n. What algorithm would you advocate for determining the sign of the permutation? You need not preserve the original contents of p.
Nice solution, Bractals! I am J ;).
A simple explanation of how it works is: Every position in the array is visited once and whenever its visited the element in this position is placed at its right position. Hence it sorts the list. Since it uses the swap operation only once per position that it visits, the total swaps are utmost n (if an element is already in its position no swap is used). Hence the complexity is linear. Cool algo. Hats off.
So, can we improve the given soln? I think yes :). Here it goes.
I = 1
SIGNUM = N%2 ? 1 : -1
while I++, I < N
if P(I) = I then
SIGNUM = -SIGNUM
Posted by And Or
on 2006-08-07 10:46:16