@all

plz.. tell if this thing would work..

assume 2 in place of every 0 in array. ie
1 1 0 0 0 1 0 1   be
1 1 2 2 2 1 2 1

then find max sub array wid sum length/2 * 3.

this can be done in O(n) but worst case would still be O(n lgn) .

On 1/26/12, Sanjay Rajpal <[email protected]> wrote:
> +1
> *
> Sanjay Kumar
> B.Tech Final Year
> Department of Computer Engineering
> National Institute of Technology Kurukshetra
> Kurukshetra - 136119
> Haryana, India
> Contact: +91-8053566286
> *
>
>
>
> On Thu, Jan 26, 2012 at 6:28 PM, Ashish Goel <[email protected]> wrote:
>
>> replace all 0s by -1
>> keep additional array to get the sumHere at every position of all -1s and
>> 1s.
>>
>> say you got
>>          0  1  0 1  0  0  0  0 1 1 1 1  0
>>         -1 1 -1 1 -1 -1 -1 -1 1 1 1 1 -1
>> sum -1 0 -1 0 -1 -2 -3 -4 -3 -2 -1 0 -1
>>
>> all equal numbers in sum shows equal zeros and 1s between then including
>> the end( between two -2 or two -3 or 0 0r -1) so biggest one can be
>> figured
>> out easily use a hash to store these cum sum, store their first and
>> last occurrence, walk over to get max diff.
>>
>>
>>
>>
>>
>> Best Regards
>> Ashish Goel
>> "Think positive and find fuel in failure"
>> +919985813081
>> +919966006652
>>
>>
>>
>> On Fri, Jan 27, 2012 at 1:48 AM, algoist <[email protected]> wrote:
>>
>>> Consider 2 temp arrays, B and C
>>>
>>> Where B gets updated for every find of 0 and C for every find of 1
>>>
>>> i.e if(a[i]==0)
>>>        b[i]+=b[i-1]+1;
>>>        c[i]=c[i-1];
>>> i.e if(a[i]==1)
>>>        c[i]+=c[i-1]+1;
>>>        b[i]=b[i-1];
>>>
>>> if(c[i]==b[i])
>>>   update max.
>>>
>>> return max.
>>>
>>> This is O(N) algo. Is it right or i am missing anything here?
>>>
>>>
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