Thread: re arrange index of numbers

One way to look at it is that you are simply sorting the array, but with your own comparison requirements.

pick up first element...if its odd then leave it...move forward...pick 2nd element...if its even then search(ahead) for the first odd number and exchange...then pick the next element...if its odd leave it....pick up next element and go on......
You'll get odd numbers in front and even numbers in the end.....

Another one could be.
take two pointers.first to the first element and second to the last.

Code:

while(first<=second)
   {       while(*first is odd)       
              { increment first;} 
         while(*second is even)
               {decrement second;}
          if (first<second)
            {  swap elements}
       }

this will do in O(n).

Do it the easy way:
1) Copy each odd number as you come across it, into a new array. Then move to step 2.
2) Copy each even number as you come across it into the new array. Then move to step 3.
3) Copy the new array over top of the first one.


Quzah.

Originally Posted by Rashakil Fol

Both sunny's and cbastard's algorithms do not solve the problem because order is not preserved.

Order was never asked by bazzano...he simply wanted to seperate odd and even numbers

The word shifted implies keeping the order. When you shift bits, you aren't reordering them. You're sliding them down one direction or another.


Quzah.

Originally Posted by quzah

The word shifted implies keeping the order. When you shift bits, you aren't reordering them. You're sliding them down one direction or another.
Quzah.

I didn't knew that


and bazzono gave this example

12 8 4 5 14 44 7 3 10

to

5 7 3 12 8 4 14 44 10

didn't ask for order

I did'nt bother about the order.
There are two ways of doing thing's.
1. brute force
2.optimal soln (or rather close to optimal because there will
always be another better soln).