http://codepad.org/RrtoXLTX

On Sun, Aug 1, 2010 at 5:09 AM, Nikhil Jindal <fundoon...@yahoo.co.in>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 difference d(for all
> possible d, which is n).
>
> Now, for it to be possible that the number of ones and number zeros between
> two indices i and j are same, a[i].diff needs to be equal to a[j].diff
>
> Thus, for a particular difference d, in the list I take the first value of
> the list and the last value and their diff gives me the maximum length of
> string possible through diff d.
>
> Now find max by iterating thru all d's which is again O(n).
>
> Hence, time complexity: O(n)
> Space: O(n)
>
> On Sun, Aug 1, 2010 at 12:33 PM, Ram Kumar <harrypotter....@gmail.com>wrote:
>
>> 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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