@Dave,
Ya. You are right. I think i overlooked your solution. The step 'set the
even position to 0...' sounded like, setting 0 and 1 directly :-)
Thanks,
Vikram
On 11/13/07, Dave <[EMAIL PROTECTED]> wrote:
>
>
> No. At step 6, we have found a 1 in an even spot and a 0 in an odd
> spot, so we
No. At step 6, we have found a 1 in an even spot and a 0 in an odd
spot, so we switch them. Switching a 1 and a 0 simply means storing 0
where the 1 was and storing 1 where the 0 was; there is no need to use
the usual code to interchange two values.
Dave
On Nov 13, 8:37 am, "Vikram Venkatesan" <
Thank you, I understood the question wrong. I thought if the # of 0's
or #1's exceeded each other, the whole array would be untouched. So i
couldn't solve it in one pass.
On 13 Kasım, 16:15, Dave <[EMAIL PROTECTED]> wrote:
> The following algorithm examine the contents of each element of the
> ar
@Dave,
>>*Otherwise, set the even position to 0 and the odd position to 1.*
I think your solution might be inserting 0's and 1's into the array from
nowhere (thus filling the whole array with alternating 0's and 1's up to the
given size !). The question is to re-arrange existing elements in th
The following algorithm examine the contents of each element of the
array at most once.
1. Start with the first even position and the first odd position.
2. Search forward through the even positions until you reach the end
of the array or find a 1, whichever comes first.
3. Search forward through
Can you explain the modification?
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A simple modification to quicksort will do the trick !
On 11/13/07, geekko <[EMAIL PROTECTED]> wrote:
>
>
> you are given an array of integers containing only 0s and 1s. You have
> to place all the 0s in even positions and 1s in odd position. And if
> suppose, no. of 0s exceeds no. of 1s or vice v
