http://codepad.org/RrtoXLTX
On Sun, Aug 1, 2010 at 5:09 AM, Nikhil Jindal wrote:
> Here's a possible O(n) soln:
>
> for all i, I calculate a[i].diff as number of zeroes - number of ones ones
> from a[0] to a[i].
>
> I do this in O(n).
>
> Also, I make a list of all the indexes that have a differe
Here's a possible O(n) soln:
for all i, I calculate a[i].diff as number of zeroes - number of ones ones
from a[0] to a[i].
I do this in O(n).
Also, I make a list of all the indexes that have a difference d(for all
possible d, which is n).
Now, for it to be possible that the number of ones and n
Given an array of 0s and 1s in any order, find the longest sequence
that has equal number of 0s and 1s.
0 0 0 0 1 1 1 1 0 0 //array
0 1 2 3 4 5 6 7 8 9 //index
ans1 (0,7)
ans2 (1,8)
ans3 (2,9)
all having 4 0's and 4 1's
--
Regards,
Ramkumar.G
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