# Thread: re arrange index of numbers

1. ## re arrange index of numbers

Hey guys just wondering how i would have an array of integers

eg

12 8 4 5 14 44 7 3 10

and rearrange it so all then odd numbers are shifted to the front
of the array and the even numbers are shifted to the end.

eg

5 7 3 12 8 4 14 44 10

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

3. 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.....

4. 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).

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

6. 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.

7. 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

8. 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.

9. 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