Question

how can i rearrange an array so all the even numbers will have indices less than those for the odd numbers.
This is in C.

I have code for the rest of the program

Answer 1

does order matters?

Answer 2

no order does not matter

Answer 3

void rearrange(int a[])
{int i, b[SIZE];
for(i=0;i<SIZE;i++)
{if(a[i]%2)
{THEN ITS ODD}

Answer 4

thats what i have so far

Answer 5

you can create new array and check old array if it's even and put those numbers there :D and later put odd numbers

Answer 6

I am not sure if i know how to do that but i can try.

if(a[i]%2)
{b[i] = a[i];} //store odd values into new array
else{} //I am not sure what to do with it if its even then. make another array?

Answer 7

no you need to index from 0, not to same position

Answer 8

and then increment when you add number
http://pastebin.com/F8HT8rmT

Answer 9

oh!!! okay that makes sense now. thanks for the help thomas, i appreciate it

Answer 10

i think it's bad solution but it works :D

Answer 11

ha I am sure there are other ways to do it but hey it works for me!

Answer 12

You could sort the numbers by their modulus value:

qsort( number % 2 );

Something like that.

Answer 13

Actually this problem is very easy. Declare another integer array with same length, and then copy even numbers first, then append remaining numbers (thee odd ones) in last of the second array. The code snippet to do this is shown below:

int[] arrange(int a[], size)
{
int *b = new int[size];
int index = 0; // variable to track position in second array
// copy even numbers first
for (int i = 0; i < size; ++i)
{
if (a[i]%2 == 0)
{
b[index] = a[i];
++b;
}
}

// copy all odd numbers
for (int i = 0; i < size; ++i)
{
if (a[i]%2 != 0)
{
b[index] = a[i];
++b;
}
}
// now return new array
return b;
}

Hope this will help you.

Happy programming.